Tangents And Normals

Given:

y = at x = 4

First, we have to find of given function, f(x),i.e, to find the derivative of f(x)

y =

⇒ y = (

⇒ y =

(xⁿ) = n.x^{n – 1}

The Slope of the tangent is

Since, x = 4

_{x = 4} =

_{x = 4} =

_{x = 4} =

_{x = 4} = 3

The Slope of the tangent at x = 4 is 3

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

Given:

y = at x = 4

First, we have to find of given function, f(x),i.e, to find the derivative of f(x)

y =

⇒ y = (

⇒ y =

(xⁿ) = n.x^{n – 1}

The Slope of the tangent is

Since, x = 4

_{x = 4} =

_{x = 4} =

_{x = 4} =

_{x = 4} = 3

The Slope of the tangent at x = 4 is 3

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

Given:

y = at x = 9

First, we have to find of given function, f(x),i.e, to find the derivative of f(x)

⇒ y =

⇒ y = (

(xⁿ) = n.x^{n – 1}

The Slope of the tangent is

⇒ y = (

Since, x = 9

_{x = 9} =

_{x = 9} =

_{x = 9} =

_{x = 9} =

_{x = 9} =

The Slope of the tangent at x = 9 is

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal = – 6

Given:

y = x³ – x at x = 2

First, we have to find of given function, f(x),i.e, to find the derivative of f(x)

(xⁿ) = n.x^{n – 1}

The Slope of the tangent is

⇒ y = x³ – x

(x³) + 3(x)

= 3.x^{3 – 1} – 1.x^{1 – 0}

= 3x² – 1

Since, x = 2

_{x = 2} = 3(2)² – 1

_{x = 2} = (34) – 1

_{x = 2} = 12 – 1

_{x = 2} = 11

The Slope of the tangent at x = 2 is 11

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

Given:

y = 2x² + 3sinx at x = 0

First, we have to find of given function, f(x),i.e, to find the derivative of f(x)

(xⁿ) = n.x^{n – 1}

The Slope of the tangent is

⇒ y = 2x² + 3sinx

⁼ 2(x²) + 3(sinx)

= 22x^{2 – 1} + 3(cosx)

(sinx) = cosx

= 4x + 3cosx

Since, x = 2

⇒ _{x = 0} = 40 + 3cos(0)

cos(0) = 1

⇒ _{x = 0} = 0 + 31

⇒ _{x = 0} = 3

The Slope of the tangent at x = 0 is 3

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

Given:

x = a() & y = a(1 + cos) at

Here, To find , we have to find & and and divide and we get our desired .

(xⁿ) = n.x^{n – 1}

⇒ x = a()

⇒ = a(() – (sin))

⇒ = a(1 – ) ...(1)

(sinx) = cosx

⇒ y = a(1 + cos)

⇒ = a(() + (cos))

(cosx) = – sinx

(Constant) = 0

⇒ = a( + ( – sin))

⇒ = a( – sin)

⇒ = – asin ...(2)

⇒

⇒

The Slope of the tangent is

Since,

⇒

sin() = 1

cos() = 0

⇒

⇒

⇒ = 1

The Slope of the tangent at x = is 1

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal = – 1

Given:

x = acos³ & y = asin³ at

Here, To find , we have to find & and and divide and we get our desired .

(xⁿ) = n.x^{n – 1}

⇒ x = acos³

⇒ = a((cos³))

(cosx) = – sinx

⇒ = a(3cos^{3 – 1} – sin)

⇒ = a(3cos² – sin)

⇒ = – 3acos²sin ...(1)

⇒ y = asin³

⇒ = a((sin³))

(sinx) = cosx

⇒ = a(3sin^{3 – 1}cos)

⇒ = a(3sin²cos)

⇒ = 3asin²cos ...(2)

⇒

⇒

⇒ = – tan

The Slope of the tangent is – tan

Since,

⇒ = – tan()

⇒ = – 1

tan() = 1

The Slope of the tangent at x = is – 1

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal = 1

Given:

x = a() & y = a(1 – cos) at

Here, To find , we have to find & and and divide and we get our desired .

(xⁿ) = n.x^{n – 1}

⇒ x = a()

⇒ = a(() – (sin))

⇒ = a(1 – ) ...(1)

(sinx) = cosx

⇒ y = a(1 – cos)

⇒ = a(() – (cos))

(cosx) = – sinx

(Constant) = 0

⇒ = a( – ( – sin))

⇒ = asin ...(2)

⇒

⇒

The Slope of the tangent is

Since,

⇒

sin() = 1

cos() = 0

⇒

⇒

⇒ = 1

The Slope of the tangent at x = is 1

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal = – 1

Given:

y = (sin2x + cotx + 2)²at x =

First, we have to find of given function, f(x),i.e, to find the derivative of f(x)

(xⁿ) = n.x^{n – 1}

The Slope of the tangent is

⇒ y = (sin2x + cotx + 2)²

= 2(sin2x + cotx + 2)^{2 – 1}(sin2x) + (cotx) + (2)}

= 2(sin2x + cotx + 2)(cos2x) + ( – cosec²x) + (0)}

(sinx) = cosx

(cotx) = – cosec²x

⇒ = 2(sin2x + cotx + 2)(2cos2x – cosec²x)

Since, x =

= 2 (sin2() + cot() + 2)(2cos2() – cosec²())

= 2 (sin() + cot() + 2) (2cos() – cosec²())

= 2 (0 + 0 + 2)(2( – 1) – 1)

sin() = 0, cos() = – 1

cot() = 0,cosec() = 1

⇒ = 2(2)( – 2 – 1)

⇒ = 4 – 3

⇒ = – 12

The Slope of the tangent at x = is – 12

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

Given:

x² + 3y + y² = 5 at (1,1)

Here we have to differentiate the above equation with respect to x.

⇒ (x² + 3y + y²) = (5)

⇒ (x²) + (3y) + (y²) = (5)

(xⁿ) = n.x^{n – 1}

⇒ 2x + 3 + 2y = 0

⇒ 2x + (3 + 2y) = 0

⇒ (3 + 2y) = – 2x

⇒

The Slope of the tangent at (1,1)is

⇒

⇒

The Slope of the tangent at (1,1) is

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

Given:

xy = 56 at (1,6)

Here we have to use the product rule for above equation.

If u and v are differentiable function, then

(UV) = U + V

(xy) = (6)

⇒ x(y) + y(x) = (5)

(Constant) = 0

⇒ x + y = 0

⇒ x = – y

⇒

The Slope of the tangent at (1,6)is

⇒

⇒ = – 6

The Slope of the tangent at (1,6) is – 6

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

⇒ The Slope of the normal =

Given:

The Slope of the tangent to the curve xy + ax + by = 2 at (1,1) is 2

First, we will find The Slope of tangent

we use product rule here,

(UV) = U + V

⇒ xy + ax + by = 2

⇒ x(y) + y(x) + a(x) + b(y) + = (2)

⇒ x + y + a + b = 0

⇒ (x + b) + y + a = 0

⇒ (x + b) = – (a + y)

⇒

since, The Slope of the tangent to the curve xy + ax + by = 2 at (1,1) is 2

i.e, = 2

⇒ {}_{(x = 1,y = 1)} = 2

⇒ = 2

⇒ – a – 1 = 2(1 + b)

⇒ – a – 1 = 2 + 2b

⇒ a + 2b = – 3 ...(1)

Also, the point (1,1) lies on the curve xy + ax + by = 2,we have

11 + a1 + b1 = 2

⇒ 1 + a + b = 2

⇒ a + b = 1 ...(2)

from (1) & (2),we get

substitute b = – 4 in a + b = 1

a – 4 = 1

⇒ a = 5

So the value of a = 5 & b = – 4

Given:

The Slope of the tangent to the curve y = x³ + ax + b at

(1, – 6)

First, we will find The Slope of tangent

y = x³ + ax + b

(x³) + (ax) + (b)

⇒ = 3x^{3 – 1} + ) +

⇒ = 3x² +

The Slope of the tangent to the curve y = x³ + ax + b at

(1, – 6) is

⇒ = 3(1)² +

⇒ = 3 + ...(1)

The given line is x – y + 5 = 0

y = x + 5 is the form of equation of a straight line y = mx + c,where m is the The Slope of the line.

so the The Slope of the line is y = 1x + 5

so The Slope is 1. ...(2)

Also the point (1, – 6) lie on the tangent, so

x = 1 & y = – 6 satisfies the equation, y = x³ + ax + b

i.e, – 6 = 1³ + a + b

⇒ – 6 = 1 + a + b

⇒ a + b = – 7 ...(3)

Since, the tangent is parallel to the line, from (1) & (2)

Hence,

3 + = 1

⇒ a = – 2

From (3)

a + b = – 7

⇒ – 2 + b = – 7

⇒ b = – 5

So the value is a = – 2 & b = – 5

Given:

The curve y = x³ – 3x

First, we will find the Slope of the tangent

y = x³ – 3x

(x³) – (3x)

⇒ = 3x^{3 – 1} – )

⇒ = 3x² – 3 ...(1)

The equation of line passing through (x₀,y₀) and The Slope m is y – y₀ = m(x – x₀).

so The Slope, m =

The Slope of the chord joining (1, – 2) & (2,2)

⇒

⇒

⇒ = 4 ...(2)

From (1) & (2)

3x² – 3 = 4

⇒ 3x² = 7

⇒ x² =

⇒ x =

y = x³ – 3x

⇒ y = x(x² – 3)

⇒ y = (()² – 3)

⇒ y = (( – 3)

⇒ y = ()

⇒ y = ()

Thus, the required point is x = & y = ()

Given:

The curve y = x³ – 2x² – 2x and a line y = 2x – 3

First, we will find The Slope of tangent

y = x³ – 2x² – 2x

(x³) – (2x²) – (2x)

⇒ = 3x^{3 – 1} – (x^{2 – 1}) – 2x^{1 – 1}

⇒ = 3x² – 4x – 2 ...(1)

y = 2x – 3 is the form of equation of a straight line y = mx + c,where m is the The Slope of the line.

so the The Slope of the line is y = 2(x) – 3

Thus, The Slope = 2. ...(2)

From (1) & (2)

⇒ 3x² – 4x – 2 = 2

⇒ 3x² – 4x = 4

⇒ 3x² – 4x – 4 = 0

We will use factorization method to solve the above Quadratic equation.

⇒ 3x² – 6x + 2x – 4 = 0

⇒ 3x(x – 2) + 2(x – 2) = 0

⇒ (x – 2)(3x + 2) = 0

⇒ (x – 2) = 0 & (3x + 2) = 0

⇒ x = 2 or

x =

Substitute x = 2 & x = in y = x³ – 2x² – 2x

when x = 2

⇒ y = (2)³ – 2 (2)² – 2(2)

⇒ y = 8 – (24) – 4

⇒ y = 8 – 8 – 4

⇒ y = – 4

when x =

⇒ y = ()³ – 2 ()² – 2()

⇒ y = () – 2 () + ()

⇒ y = () – () + ()

taking lcm

⇒ y =

⇒ y =

⇒ y =

Thus, the points are (2, – 4) & (,)

Given:

The curve y² = 2x³ and The Slope of tangent is 3

y² = 2x³

Differentiating the above w.r.t x

⇒ 2y^{2 – 1} = 2x^{3 – 1}

⇒ y = 3x²

⇒

Since, The Slope of tangent is 3

= 3

⇒ = 1

⇒ x² = y

Substituting x² = y in y² = 2x³,

(x²)² = 2x³

x⁴ – 2x³ = 0

x³(x – 2) = 0

x³ = 0 or (x – 2) = 0

x = 0 or x = 2

If x = 0

⇒

⇒ , which is not possible.

So we take x = 2 and substitute it in y² = 2x³,we get

y² = 2(2)³

y² = 28

y² = 16

y = 4

Thus, the required point is (2,4)

Given:

The curve is xy + 4 = 0

If a tangent line to the curve y = f(x) makes an angle with x – axis in the positive direction, then

= The Slope of the tangent = tan

xy + 4 = 0

Differentiating the above w.r.t x

⇒ x(y) + y(x) + (4) = 0

⇒ x + y = 0

⇒ x = – y

⇒ ...(1)

Also, = tan45° = 1 ...(2)

From (1) & (2),we get,

⇒ = 1

⇒ x = – y

Substitute in xy + 4 = 0,we get

⇒ x( – x) + 4 = 0

⇒ – x² + 4 = 0

⇒ x² = 4

⇒ x = 2

so when x = 2,y = – 2

& when x = – 2,y = 2

Thus, the points are (2, – 2) & ( – 2,2)

Given:

The curve is y = x²

y = x²

Differentiating the above w.r.t x

⇒ = 2x^{2 – 1}

⇒ = 2x ...(1)

Also given the Slope of the tangent is equal to the x – coordinate,

= x ...(2)

From (1) & (2),we get,

i.e,2x = x

⇒ x = 0.

Substituting this in y = x², we get,

y = 0²

⇒ y = 0

Thus, the required point is (0,0)

Given:

The curve is x² + y² – 2x – 4y + 1 = 0

Differentiating the above w.r.t x

⇒ x² + y² – 2x – 4y + 1 = 0

⇒ 2x^{2 – 1} + 2y^{2 – 1} – 2 – 4 + 0 = 0

⇒ 2x + 2y – 2 – 4 = 0

⇒ (2y – 4) = – 2x + 2

⇒

...(1)

= The Slope of the tangent = tan

Since, the tangent is parallel to x – axis

i.e,

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

From (1) & (2),we get,

⇒ = 0

⇒ – (x – 1) = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 4y + 1 = 0,we get,

⇒ 1² + y² – 21 – 4y + 1 = 0

⇒ 1 – y² – 2 – 4y + 1 = 0

⇒ y² – 4y = 0

⇒ y(y – 4) = 0

⇒ y = 0 & y = 4

Thus, the required point is (1,0) & (1,4)

Given:

The curve is y = x²

Differentiating the above w.r.t x

⇒ y = x²

⇒ = 2x^{2 – 1}

⇒ = 2x ...(1)

= The Slope of the tangent = tan

Since, the tangent make an angle of 45^o with x – axis

i.e,

⇒ = tan(45°) = 1 ...(2)

tan(45°) = 1

From (1) & (2), we get,

⇒ 2x = 1

⇒ x =

Substituting x = in y = x², we get,

⇒ y = ()²

⇒ y =

Thus, the required point is (,)

Given:

The curve is y = 3x² – 9x + 8

Differentiating the above w.r.t x

⇒ y = 3x² – 9x + 8

⇒ = 23x^{2 – 1} – 9 + 0

⇒ = 6x – 9 ...(1)

Since, the tangent are equally inclined with axes

i.e, or

= The Slope of the tangent = tan

⇒ = tan() or tan()

⇒ = 1or – 1 ...(2)

tan() = 1

From (1) & (2),we get,

⇒ 6x – 9 = 1 0r 6x – 9 = – 1

⇒ 6x = 10 0r 6x = 8

⇒ x = or x =

⇒ x = or x =

Substituting x = or x = in y = 3x² – 9x + 8,we get,

When x =

⇒ y = 3()² – 9() + 8

⇒ y = 3() – () + 8

⇒ y = () – () + 8

taking LCM = 9

⇒ y = ()

⇒ y = ()

⇒ y = ()

⇒ y = ()

when x =

⇒ y = 3()² – 9() + 8

⇒ y = 3() – () + 8

⇒ y = () – () + 8

taking LCM = 9

⇒ y = ()

⇒ y = ()

⇒ y = ()

⇒ y = ()

Thus, the required point is (,) & (,)

Given:

The curve is y = 2x² – x + 1and the line y = 3x + 4

First, we will find The Slope of tangent

y = 2x² – x + 1

⇒ (2x²) – (x) + (1)

⇒ = 4x – 1 ...(1)

y = 3x + 4 is the form of equation of a straight line y = mx + c,where m is the The Slope of the line.

so the The Slope of the line is y = 3(x) + 4

Thus, The Slope = 3. ...(2)

From (1) & (2),we get,

4x – 1 = 3

⇒ 4x = 4

⇒ x = 1

Substituting x = 1in y = 2x² – x + 1,we get,

⇒ y = 2(1)² – (1) + 1

⇒ y = 2 – 1 + 1

⇒ y = 2

Thus, the required point is (1,2)

Given:

The curve y = 3x² + 4 and the Slope of the tangent is

y = 3x² + 4

Differentiating the above w.r.t x

⇒ = 23x^{2 – 1} + 0

⇒ = 6x ...(1)

Since, tangent is perpendicular to the line,

The Slope of the normal =

i.e, =

⇒ =

⇒ x = 1

Substituting x = 1 in y = 3x² + 4,

⇒ y = 3(1)² + 4

⇒ y = 3 + 4

⇒ y = 7

Thus, the required point is (1,7).

Given:

The curve x² + y² = 13 and the line 2x + 3y = 7

x² + y² = 13

Differentiating the above w.r.t x

⇒ 2x^{2 – 1} + 2y^{2 – 1} = 0

⇒ 2x + 2y = 0

⇒ 2(x + y) = 0

⇒ (x + y) = 0

⇒ y = – x

⇒ ...(1)

Since, line is 2x + 3y = 7

⇒ 3y = – 2x + 7

⇒ y =

⇒ y = +

The equation of a straight line is y = mx + c, where m is the The Slope of the line.

Thus, the The Slope of the line is ...(2)

Since, tangent is parallel to the line,

the The Slope of the tangent = The Slope of the normal

=

⇒ – x =

⇒ x =

Substituting x = in x² + y² = 13,

⇒ ()² + y² = 13

⇒ () + y² = 13

⇒ y²() = 13

⇒ y²() = 13

⇒ y²() = 1

⇒ y² = 9

⇒ y = 3

Substituting y = 3 in x = ,we get,

x =

x = 2

Thus, the required point is (2, 3) & ( – 2, – 3)

Given:

The curve is 2a²y = x³ – 3ax²

Differentiating the above w.r.t x

⇒ 2a² = 3x^{3 – 1} – 32ax^{2 – 1}

⇒ 2a² = 3x² – 6ax

⇒ = ...(1)

= The Slope of the tangent = tan

Since, the tangent is parallel to x – axis

i.e,

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

= The Slope of the tangent = tan

From (1) & (2),we get,

⇒ = 0

⇒ 3x² – 6ax = 0

⇒ 3x(x – 2a) = 0

⇒ 3x = 0 or (x – 2a) = 0

⇒ x = 0 or x = 2a

Substituting x = 0 or x = 2a in 2a²y = x³ – 3ax²,

when x = 0

⇒ 2a²y = (0)³ – 3a(0)²

⇒ y = 0

when x = 2

⇒ 2a²y = (2a)³ – 3a(2a)²

⇒ 2a²y = 8a³ – 12a³

⇒ 2a²y = – 4a³

⇒ y = – 2a

Thus, the required point is (0,0) & (2a, – 2a)

Given:

The curve y = x² – 4x + 5 and line is 2y + x = 7

y = x² – 4x + 5

Differentiating the above w.r.t x,

we get the Slope of the tangent,

⇒ = 2x^{2 – 1} – 4 + 0

⇒ = 2x – 4 ...(1)

Since, line is 2y + x = 7

⇒ 2y = – x + 7

⇒ y =

⇒ y = +

The equation of a straight line is y = mx + c, where m is the The Slope of the line.

Thus, the The Slope of the line is ...(2)

Since, tangent is perpendicular to the line,

The Slope of the normal =

From (1) & (2),we get

i.e, =

⇒ 1 =

⇒ x – 2 = 1

⇒ x = 3

Substituting x = 3 in y = x² – 4x + 5,

⇒ y = y = 3² – 4×3 + 5

⇒ y = 9 – 12 + 5

⇒ y = 2

Thus, the required point is (3,2)

Given:

The curve is = 1

Differentiating the above w.r.t x, we get the The Slope of a tangent,

⇒ = 0

Cross multiplying we get,

⇒ = 0

⇒ 50x + 8y = 0

⇒ 8y = – 50x

⇒ =

⇒ = ...(1)

(i)

Since, the tangent is parallel to x – axis

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

= The Slope of the tangent = tan

From (1) & (2),we get,

⇒ = 0

⇒ – 25x = 0

⇒ x = 0

Substituting x = 0 in = 1,

= 1

⇒ y² = 25

⇒ y = ±5

Thus, the required point is (0,5) & (0, – 5)

Since, the tangent is parallel to y – axis, its The Slope is not defined, then the normal is parallel to x – axis whose The Slope is zero.

i.e, = 0

⇒ = 0

⇒ = 0

⇒ y = 0

Substituting y = 0 in = 1,

= 1

⇒ x² = 4

⇒ x = ±2

Thus, the required point is (2,0) & ( – 2,0)

Given:

The curve is x² + y² – 2x – 3 = 0

Differentiating the above w.r.t x, we get The Slope of tangent,

⇒ 2x^{2 – 1} + 2y^{2 – 1} – 2 – 0 = 0

⇒ 2x + 2y – 2 = 0

⇒ 2y = 2 – 2x

⇒ =

⇒ = ...(1)

(i) Since, the tangent is parallel to x – axis

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

= The Slope of the tangent = tan

From (1) & (2),we get,

⇒ = 0

⇒ 1 – x = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 3 = 0,

1² + y² – 2×1 – 3 = 0

1 + y² – 2 – 3 = 0

y² – 4 = 0

⇒ y² = 4

⇒ y = ±2

Thus, the required point is (1,2) & (1, – 2)

Since, the tangent is parallel to y – axis, its slope is not defined, then the normal is parallel to x – axis whose slope is zero.

i.e, = 0

⇒ = 0

⇒ = 0

⇒ y = 0

Substituting y = 0 in x² + y² – 2x – 3 = 0,

x² + 0² – 2×x – 3 = 0

x² – 2x – 3 = 0

Using factorization method, we can solve above quadratic equation

x² – 3x + x – 3 = 0

x(x – 3) + 1(x – 3) = 0

(x – 3)(x + 1) = 0

⇒ x = 3 & x = – 1

Thus, the required point is (3,0) & ( – 1,0)

Given:

The curve is = 1

Differentiating the above w.r.t x, we get the Slope of tangent,

⇒ = 0

⇒ = 0

Cross multiplying we get,

⇒ = 0

⇒ 16x + 9y = 0

⇒ 9y = – 16x

⇒ = ...(1)

(i)

Since, the tangent is parallel to x – axis

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

= The Slope of the tangent = tan

From (1) & (2),we get,

⇒ = 0

⇒ – 16x = 0

⇒ x = 0

Substituting x = 0 in = 1,

= 1

⇒ y² = 16

⇒ y = ±4

Thus, the required point is (0,4) & (0, – 4)

Since the tangent is parallel to y–axis, its slope is not defined, then the normal is parallel to x–axis whose The Slope is zero.

i.e., = 0

⇒ = 0

⇒ = 0

⇒ y = 0

Substituting y = 0 in = 1,

= 1

⇒ x² = 9

⇒ x = ±3

Thus, the required point is (3,0) & ( – 3,0)

Given:

The curve y = 7x³ + 11

Differentiating the above w.r.t x

⇒ = 37x^{3 – 1} + 0

⇒ = 21x²

when x = 2

⇒ = 21×(2)²

⇒ = 21×4

⇒ = 84

when x = – 2

⇒ = 21×( – 2)²

⇒ = 21×4

⇒ = 84

Let y = f(x) be a continuous function and P(x_0,y₀) be point on the curve, then,

The Slope of the tangent at P(x_,y) is f'(x) or

Since, the Slope of the tangent is at x = 2 and x = – 2 are equal, the tangents at x = 2 and x = – 2 are parallel.

Given:

The curve is y = x³

y = x³

Differentiating the above w.r.t x

⇒ = 3x^{2 – 1}

⇒ = 3x² ...(1)

Also given the The Slope of the tangent is equal to the x – coordinate,

= x ...(2)

From (1) & (2),we get,

i.e, 3x² = x

⇒ x(3x – 1) = 0

⇒ x = 0 or x =

Substituting x = 0 or x = this in y = x³,we get,

when x = 0

⇒ y = 0³

⇒ y = 0

when x =

⇒ y = )³

⇒ y =

Thus, the required point is (0,0) & (,)

finding slope of the tangent by differentiating the curve

at slope m, is – 1

the equation of the tangent is given by y – y₁ = m(x – x₁)

finding the slope of the tangent by differentiating the curve

m = 4 at (1,4)

normal is perpendicular to tangent so, m₁m₂ = – 1

equation of normal is given by y – y₁ = m(normal)(x – x₁)

x + 4y = 17

finding the slope of the tangent by differentiating the curve

m(tangent) at (0,5) = – 10

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 5 = – 10x

y + 10x = 5

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating the curve

m(tangent) at (x = 1) = 2

normal is perpendicular to tangent so, m₁m₂ = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 3 = 2(x – 1)

y = 2x + 1

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

m(tangent) at (x = 0) = 0

normal is perpendicular to tangent so, m₁m₂ = – 1

We can see that the slope of normal is not defined

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y = 0

equation of normal is given by y – y₁ = m(normal)(x – x₁)

x = 0

finding the slope of the tangent by differentiating the curve

m(tangent) at (1, – 2) = 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (1, – 2) = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y + 2 = 1(x – 1)

y = x – 3

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y + 2 = – 1(x – 1)

y + x + 1 = 0

finding the slope of the tangent by differentiating the curve

m(tangent) at (2, – 2) = – 2

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y + 2 = – 2(x – 2)

y + 2x = 2

equation of normal is given by y – y₁ = m(normal)(x – x₁)

2y + 4 = x – 2

2y – x + 6 = 0

finding slope of the tangent by differentiating the curve

m(tangent) at (3,0) = 10

normal is perpendicular to tangent so, m₁m₂ = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y at x = 3

y = 3² + 4×3 + 1

y = 22

y – 22 = 10(x – 3)

y = 10x – 8

equation of normal is given by y – y₁ = m(normal)(x – x₁)

x + 10y = 223

finding the slope of the tangent by differentiating the curve

normal is perpendicular to tangent so, m₁m₂ = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

normal is perpendicular to tangent so, m₁m₂ = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

m(tangent) at

m(tangent) = m

normal is perpendicular to tangent so, m₁m₂ = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

m(tangent) at =

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating the curve

m(tangent) at =

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

m(tangent) at (x₁, y₁) =

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (x₁, y₁) =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

m(tangent) at (x₀, y₀) =

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (x₁, y₁) =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

m(tangent) at (1,1) = – 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (1,1) = 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 1 = – 1(x – 1)

x + y = 2

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y – 1 = 1(x – 1)

y = x

finding the slope of the tangent by differentiating the curve

m(tangent) at (2,1) = 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (2,1) = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 1 = 1(x – 2)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y – 1 = – 1(x – 2)

finding slope of the tangent by differentiating the curve

m(tangent) at (x₁, y₁) =

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (x₁, y₁) =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

m(tangent) at (1,2) = 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (1,2) = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 2 = 1(x – 1)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y – 2 = – 1(x – 1)

finding the slope of the tangent by differentiating the curve

m(tangent) at (3 cos θ, 2 sin θ) =

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (3 cos θ, 2 sin θ) =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating the curve

normal is perpendicular to tangent so, m₁m₂ = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating x and y with respect to theta

Dividing both the above equations

m at theta ( ) =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

finding slope of the tangent by differentiating x and y with respect to theta

Dividing both the above equations

m(tangent) at theta ( ) = – 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at theta ( ) = 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating x and y with respect to t

Now dividing and to obtain the slope of tangent

m(tangent) at t = is

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at t = is

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating x and y with respect to t

Now dividing and to obtain the slope of tangent

m(tangent) at t = 1 is 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at t = 1 is – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 2a = 1(x – a)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y – 2a = – 1(x – a)

finding slope of the tangent by differentiating x and y with respect to t

Now dividing and to obtain the slope of tangent

m(tangent) at t =

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at t =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating x and y with respect to theta

Now dividing and to obtain the slope of tangent

m(tangent) at theta is

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at theta is

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating x and y with respect to theta

Now dividing and to obtain the slope of tangent

m(tangent) at theta is

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at theta is

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding slope of the tangent by differentiating the curve

Finding y co – ordinate by substituting x in the given curve

2y² – 6y + 4 = 0

y² – 3y + 2 = 0

y = 2 or y = 1

m(tangent) at x = 2 is 0

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at x = 2 is , which is undefined

equation of normal is given by y – y₁ = m(normal)(x – x₁)

x = 2

finding the slope of the tangent by differentiating the curve

m(tangent) at (am², am³) is

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (am², am³) is

equation of normal is given by y – y₁ = m(normal)(x – x₁)

finding the slope of the tangent by differentiating the curve

m(tangent) at (2,3) = 2a

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

2a = 4

a = 2

now (2,3) lies on the curve, these points must satisfy

3² = 2 × 2³ + b

b = – 7

finding the slope of the tangent by differentiating the curve

m(tangent) = 2x + 4

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

2x + 4 = 3

Now substituting the value of x in the curve to find y

Therefore, the equation of tangent parallel to the given line is

finding the slope of the tangent by differentiating the curve

m(tangent) = 3x² + 2

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) =

equation of normal is given by y – y₁ = m(normal)(x – x₁)

now comparing the slope of normal with the given equation

m(normal) =

x = 2 or – 2

hence the corresponding value of y is 18 or – 6

so, equations of normal are

Or

finding the slope of the tangent by differentiating the curve

m(tangent) =

the slope of given line is , so the slope of line perpendicular to it is 9

x = 1 or – 1

since this point lies on the curve, we can find y by substituting x

y = 6 or 4

therefore, the equation of the tangent is given by

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 6 = 9(x – 1)

or

y – 4 = 9(x + 1)

finding the slope of the tangent by differentiating the curve

m(tangent) =

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) =

equation of normal is given by y – y₁ = m(normal)(x – x₁)

now comparing the slope of normal with the given equation

m(normal) = 1

since this point lies on the curve, we can find y by substituting x

The equation of normal is given by

finding the slope of the tangent by differentiating the curve

m(tangent) = 2x – 2

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

m(tangent) = 2

2x – 2 = 2

x = 2

since this point lies on the curve, we can find y by substituting x

y = 2² – 2 × 2 + 7

y = 7

therefore, the equation of the tangent is

y – 7 = 2(x – 2)

slope of given line is 3

finding the slope of the tangent by differentiating the curve

m(tangent) = 2x – 2

since both lines are perpendicular to each other

(2x – 2) × 3 = – 1

since this point lies on the curve, we can find y by substituting x

therefore, the equation of the tangent is

finding the slope of the tangent by differentiating the curve

Now according to question, the slope of all tangents is equal to 2, so

We can see that LHS is always greater than or equal to 0, while RHS is always negative. Hence no tangent is possible

finding the slope of the tangent by differentiating the curve

Now according to question, the slope of all tangents is equal to 0, so

Therefore the only possible solution is x = 1

since this point lies on the curve, we can find y by substituting x

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

finding the slope of the tangent by differentiating the curve

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

m(tangent) = 2

since this point lies on the curve, we can find y by substituting x

therefore, the equation of the tangent is

finding the slope of the tangent by differentiating the curve

m(tangent) =

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

m(tangent) = 4

x = – 6

since this point lies on the curve, we can find y by substituting x

6² + 3y – 3 = 0

y = – 11

therefore, the equation of the tangent is

y + 11 = 4(x + 6)

finding the slope of the tangent by differentiating the curve

m(tangent) at (a,b) is

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

therefore, the equation of the tangent is

Hence, proved

finding the slope of the tangent by differentiating x and y with respect to t

Dividing the above equations to obtain the slope of the given tangent

m(tangent) at is

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

therefore, equation of tangent is

finding the slope of the tangent by differentiating the curve

According to the question, tangent is parallel to the x – axis , which implies m = 0

x = 3 or x = 2

since this point lies on the curve, we can find y by substituting x

y = 2(3)³ – 15(3)² + 36(3) – 21

y = 6

or

y = 2(2)³ – 15(2)² + 36(2) – 21

y = 7

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 6 = 0(x – 3)

y = 6

or

y – 7 = 0(x – 2)

y = 7

assume point (a, b) which lies on the given curve

finding the slope of the tangent by differentiating the curve

m(tangent) at (a,b) is

Since this tangent passes through , its slope can also be written as

Equating both the slopes as they are of the same tangent

b² = 3a² – 4a …(i)

Since points (a,b) lies on this curve

3a² – b² = 8 …(ii)

Solving (i) and (ii) we get

3a² – 8 = 3a² – 4a

a = 2

b = 2 or – 2

therefore points are (2,2) or (2, – 2)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 2 = 3(x – 2)

or

y + 2 = – 3(x – 3)

Given:

Curves y² = x ...(1)

& x² = y ...(2)

First curve is y² = x

Differentiating above w.r.t x,

⇒ 2y. = 1

⇒ m₁ ...(3)

The second curve is x² = y

⇒ 2x

⇒ m₂ = 2x ...(4)

Substituting (1) in (2),we get

⇒ x² = y

⇒ (y²)² = y

⇒ y⁴ – y = 0

⇒ y(y³ – 1) = 0

⇒ y = 0 or y = 1

Substituting y = 0 & y = 1 in (1) in (2),

x = y²

when y = 0,x = 0

when y = 1,x = 1

Substituting above values for m₁ & m_2,we get,

when x = 0,

m₁∞

when x = 1,

m₁

Values of m₁ is ∞ &

when y = 0,

m₂ = 2x = 2×0 = 0

when x = 1,

m₂ = 3x = 2×1 = 2

Values of m₂ is 0 & 2

when m₁ = ∞ & m₂ = 0

tanθ

tanθ

tanθ = ∞

θ = tan ^{– 1}(∞)

∴ tan ^{– 1}(∞)

θ

when m₁ & m₂ = 2

tanθ

tanθ

tanθ

θ = tan ^{– 1}()

θ≅36.86

Given:

Curves y = x² ...(1)

& x² + y² = 20 ...(2)

First curve y = x²

⇒ m₁ = 2x ...(3)

Second curve is x² + y² = 20

Differentiating above w.r.t x,

⇒ 2x + 2y. = 0

⇒ y. = – x

⇒ m₂ ...(4)

Substituting (1) in (2),we get

⇒ y + y² = 20

⇒ y² + y – 20 = 0

We will use factorization method to solve the above Quadratic equation

⇒ y² + 5y – 4y – 20 = 0

⇒ y(y + 5) – 4(y + 5) = 0

⇒ (y + 5)(y – 4) = 0

⇒ y = – 5 & y = 4

Substituting y = – 5 & y = 4 in (1) in (2),

y = x²

when y = – 5,

⇒ – 5 = x²

⇒ x

when y = 4,

⇒ 4 = x²

⇒ x = ±2

Substituting above values for m₁ & m_2,we get,

when x = 2,

m₁4

when x = 1,

m₁4

Values of m₁ is 4 & – 4

when y = 4 & x = 2

m₂

when y = 4 & x = – 2

m₂

Values of m₂ is &

when m₁ = ∞ & m₂ = 0

tanθ

tanθ

tanθ

θ = tan ^{– 1}()

θ≅77.47

Given:

Curves 2y² = x³ ...(1)

& y² = 32x ...(2)

First curve is 2y² = x³

Differentiating above w.r.t x,

⇒ 4y. = 3x²

⇒ m₁ ...(3)

Second curve is y² = 32x

⇒ 2y. = 32

⇒ y. = 16

⇒ m₂ ...(4)

Substituting (2) in (1),we get

⇒ 2y² = x³

⇒ 2(32x) = x³

⇒ 64x = x³

⇒ x³ – 64x = 0

⇒ x(x² – 64) = 0

⇒ x = 0 & (x² – 64) = 0

⇒ x = 0 & ±8

Substituting x = 0 & x = ±8 in (1) in (2),

y² = 32x

when x = 0,y = 0

when x = 8

⇒ y² = 32×8

⇒ y² = 256

⇒ y = ±16

Substituting above values for m₁ & m_2,we get,

when x = 0,y = 16

m₁

⇒ 0

when x = 8,y = 16

m₁

⇒ 3

Values of m₁ is 0 & 3

when x = 0,y = 0,

m₂

⇒ ∞

when y = 16,

m₂

⇒ 1

Values of m₂ is ∞ & 1

when m₁ = 0 & m₂ = ∞

⇒ tanθ

⇒ tanθ

⇒ tanθ = ∞

⇒ θ = tan ^{– 1}(∞)

∴ tan ^{– 1}(∞)

⇒ θ

when m₁ & m₂ = 2

⇒ tanθ

⇒ tanθ

⇒ tanθ

⇒ θ = tan ^{– 1}()

⇒ θ≅25.516

Given:

Curves x² + y² – 4x – 1 = 0 ...(1)

& x² + y² – 2y – 9 = 0 ...(2)

First curve is x² + y² – 4x – 1 = 0

⇒ x² – 4x + 4 + y² – 4 – 1 = 0

⇒ (x – 2)² + y² – 5 = 0

Now ,Subtracting (2) from (1),we get

⇒ x² + y² – 4x – 1 – ( x² + y² – 2y – 9) = 0

⇒ x² + y² – 4x – 1 – x² – y² + 2y + 9 = 0

⇒ – 4x – 1 + 2y + 9 = 0

⇒ – 4x + 2y + 8 = 0

⇒ 2y = 4x – 8

⇒ y = 2x – 4

Substituting y = 2x – 4 in (3),we get,

⇒ (x – 2)² + (2x – 4)² – 5 = 0

⇒ (x – 2)² + 4(x – 2)² – 5 = 0

⇒ (x – 2)²(1 + 4) – 5 = 0

⇒ 5(x – 2)² – 5 = 0

⇒ (x – 2)² – 1 = 0

⇒ (x – 2)² = 1

⇒ (x – 2) = ±1

⇒ x = 1 + 2 or x = – 1 + 2

⇒ x = 3 or x = 1

So ,when x = 3

y = 2×3 – 4

⇒ y = 6 – 4 = 2

So ,when x = 3

y = 2×1 – 4

⇒ y = 2 – 4 = – 2

The point of intersection of two curves are (3,2) & (1, – 2)

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ x² + y² – 4x – 1 = 0

⇒ 2x + 2y.4 – 0 = 0

⇒ x + y.2 = 0

⇒ y. = 2 – x

...(3)

⇒ x² + y² – 2y – 9 = 0

⇒ 2x + 2y.20 = 0

⇒ x + y.0

⇒ x + (y – 1)0

⇒ ...(4)

At (3,2) in equation(3),we get

m₁

At (3,2) in equation(4),we get

= – 3

m₂ = – 3

when m₁ & m₂ = 0

tanθ

tanθ

tanθ

tanθ = 7

θ = tan ^{– 1}(7)

θ≅81.86

Given:

Curves + 1 ...(1)

& x² + y² = ab ...(2)

Second curve is x² + y² = ab

y² = ab – x²

Substituting this in equation (1),

+ 1

1

x²b² + a³b – a²x² = a²b²

x²b² – a²x² = a²b² – a³b

x²(b² – a²) = a²b(b – a)

x²

x²

x²

∴a² – b² = (a + b)(a – b)

x ...(3)

since , y² = ab – x²

y² = ab – ()

y²

y²

y = ± ...(4)

since ,curves are + 1 & x² + y² = ab

Differentiating above w.r.t x,

⇒ . = 0

⇒ . =

⇒

⇒

⇒ m₁ ...(5)

Second curve is x² + y² = ab

⇒ 2x + 2y.0

⇒ m₂ ...(6)

Substituting (3) in (4), above values for m₁ & m_2,we get,

At (, ) in equation(5),we get

⇒ m₁

At (, ) in equation(6),we get

m₂

when m₁ & m₂

tanθ

tanθ

tanθ

tanθ

tanθ

tanθ

tanθ

θ = tan ^{– 1}()

Given:

Curves x² + 4y² = 8 ...(1)

& x² – 2y² = 2 ...(2)

Solving (1) & (2),we get,

from 2nd curve,

x² = 2 + 2y²

Substituting on x² + 4y² = 8,

⇒ 2 + 2y² + 4y² = 8

⇒ 6y² = 6

⇒ y² = 1

⇒ y = ±1

Substituting on y = ±1,we get,

⇒ x² = 2 + 2(±1)²

⇒ x² = 4

⇒ x = ±2

∴ The point of intersection of two curves (2,1) & ( – 2, – 1)

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ x² + 4y² = 8

⇒ 2x + 8y. = 0

⇒ 8y. = – 2x

...(3)

⇒ x² – 2y² = 2

⇒ 2x – 4y.0

⇒ x – 2y.0

⇒ 4yx

⇒ ...(4)

At (2,1) in equation(3),we get

m₁

At (2,1) in equation(4),we get

= 1

m₂ = 1

when m₁ & m₂ = 1

tanθ

tanθ

tanθ

tanθ

θ = tan ^{– 1}(3)

θ≅71.56

Given:

Curves x² = 27y ...(1)

& y² = 8x ...(2)

Solving (1) & (2),we get,

From y² = 8x,we get,

⇒ x

Substituting x on x² = 27y ,

⇒ ()² = 27y

⇒ () = 27y

⇒ y⁴ = 1728y

⇒ y(y³ – 1728) = 0

⇒ y = 0 or (y³ – 1728) = 0

⇒ y = 0 or y

∴

⇒ y = 0 or y = 12

Substituting y = 0 or y = 12 on x

when y = 0,

⇒ x

⇒ x = 0

when y = 12,

⇒ x

⇒ x = 18

∴ The point of intersection of two curves (0,0) & (18,12)

First curve is x² = 27y

Differentiating above w.r.t x,

⇒ 2x= 27.

⇒

⇒ m₁ ...(3)

Second curve is y² = 8x

⇒ 2y. = 8

⇒ y. = 4

⇒ m₂ ...(4)

Substituting (18,12) for m₁ & m_2,we get,

m₁

⇒

m₁ ...(5)

m₂

⇒

m₂ ...(6)

when m₁ & m₂

⇒ tanθ

⇒ tanθ

⇒ tanθ

⇒ tanθ

⇒ θ = tan ^{– 1}()

⇒ θ≅34.69

Given:

Curves x² + y² = 2x ...(1)

& y² = x ...(2)

Solving (1) & (2),we get

Substituting y² = x in x² + y² = 2x

⇒ x² + x = 2x

⇒ x² – x = 0

⇒ x(x – 1) = 0

⇒ x = 0 or (x – 1) = 0

⇒ x = 0 or x = 1

Substituting x = 0 or x = 1in y² = x ,we get,

when x = 0,

⇒ y² = 0

⇒ y = 0

when x = 1,

⇒ y² = 1

⇒ y = 1

The point of intersection of two curves are (0,0) & (1,1)

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ x² + y² = 2x

⇒ 2x + 2y. = 2

⇒ x + y. = 1

⇒ y. = 1 – x

...(3)

⇒ y² = x

⇒ 2y.1

⇒ ...(4)

At (1,1) in equation(3),we get

m₁ = 0

At (1,1) in equation(4),we get

m₂

when m₁ = 0 & m₂

tanθ

tanθ

tanθ

tanθ

θ = tan ^{– 1}()

θ≅26.56

Given:

Curves y = 4 – x² ...(1)

& y = x² ...(2)

Solving (1) & (2),we get

⇒ y = 4 – x²

⇒ x² = 4 – x²

⇒ 2x² = 4

⇒ x² = 2

⇒ x = ±

Substituting in y = x² ,we get

y = ()²

y = 2

The point of intersection of two curves are (,2) & (, – 2)

First curve y = 4 – x²

Differentiating above w.r.t x,

⇒ = 0 – 2x

⇒ m₁ = – 2x ...(3)

Second curve y = x²

Differentiating above w.r.t x,

⇒ = 2x

m₂ = 2x ...(4)

At (,2),we have,

m₁ = – 2x

⇒ – 2×

⇒ m₁ = – 2

At (,2),we have,

m₂ = – 2x

2 = 2

When m_{1 = –} 2 & m₂ = 2

tanθ

tanθ

tanθ

tanθ

θ = tan ^{– 1}()

θ≅38.94

Given:

Curves y = x³ ...(1)

& 6y = 7 – x² ...(2)

Solving (1) & (2),we get

⇒ 6y = 7 – x²

⇒ 6(x³) = 7 – x²

⇒ 6x³ + x² – 7 = 0

Since f(x) = 6x³ + x² – 7,

we have to find f(x) = 0,so that x is a factor of f(x).

when x = 1

f(1) = 6(1)³ + (1)² – 7

f(1) = 6 + 1 – 7

f(1) = 0

Hence, x = 1 is a factor of f(x).

Substituting x = 1 in y = x³ ,we get

y = 1³

y = 1

The point of intersection of two curves is (1,1)

First curve y = x³

Differentiating above w.r.t x,

⇒ m₁ = 3x² ...(3)

Second curve 6y = 7 – x²

Differentiating above w.r.t x,

⇒ 6 = 0 – 2x

⇒ m₂

⇒ m₂ ...(4)

At (1,1),we have,

m₁ = 3x²

⇒ 3×(1)²

m₁ = 3

At (1,1),we have,

⇒ m₂

⇒

⇒ m₂

When m_{1 =} 3 & m₂ =

⇒ 3×1

∴ Two curves y = x³ & 6y = 7 – x² intersect orthogonally.

Given:

Curves x³ – 3xy² = – 2 ...(1)

& 3x²y – y³ = 2 ...(2)

Adding (1) & (2),we get

⇒ x³ – 3xy² + 3x²y – y³ = – 2 + 2

⇒ x³ – 3xy² + 3x²y – y³ = – 0

⇒ (x – y)³ = 0

⇒ (x – y) = 0

⇒ x = y

Substituting x = y on x³ – 3xy² = – 2

⇒ x³ – 3×x×x² = – 2

⇒ x³ – 3x³ = – 2

⇒ – 2x³ = – 2

⇒ x³ = 1

⇒ x = 1

Since x = y

y = 1

The point of intersection of two curves is (1,1)

First curve x³ – 3xy² = – 2

Differentiating above w.r.t x,

⇒ 3x² – 3(1×y² + x×2y) = 0

⇒ 3x² – 3y² – 6xy0

⇒ 3x² – 3y² = 6xy

⇒

⇒

⇒ m₁ ...(3)

Second curve 3x²y – y³ = 2

Differentiating above w.r.t x,

⇒ 3(2x×y + x²×) – 3y²0

⇒ 6xy + 3x²3y²0

⇒ 6xy + (3x² – 3y²)0

⇒

⇒

⇒ m₂ ...(4)

When m₁ & m₂ =

⇒ × = – 1

∴ Two curves x³ – 3xy² = – 2 & 3x²y – y³ = 2 intersect orthogonally.

Given:

Curves x² + 4y² = 8 ...(1)

& x² – 2y² = 4 ...(2)

Solving (1) & (2),we get,

from 2nd curve,

x² = 4 + 2y²

Substituting on x² + 4y² = 8,

⇒ 4 + 2y² + 4y² = 8

⇒ 6y² = 4

⇒ y²

⇒ y = ±

Substituting on y = ±, we get,

⇒ x² = 4 + 2(±)²

⇒ x² = 4 + 2()

⇒ x² = 4 +

⇒ x²

⇒ x = ±

⇒ x = ±

∴ The point of intersection of two curves (,) & (,)

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ x² + 4y² = 8

⇒ 2x + 8y. = 0

⇒ 8y. = – 2x

...(3)

⇒ x² – 2y² = 4

⇒ 2x – 4y.0

⇒ x – 2y.0

⇒ 4yx

⇒ ...(4)

At (,) in equation(3),we get

m₁

At (,) in equation(4),we get

m₂ = 1

when m₁ & m₂

⇒ × = – 1

∴ Two curves x² + 4y² = 8 & x² – 2y² = 4 intersect orthogonally.

Given:

Curves x² = 4y ...(1)

& 4y + x² = 8 ...(2)

The point of intersection of two curves (2,1)

Solving (1) & (2),we get,

First curve is x² = 4y

Differentiating above w.r.t x,

⇒ 2x= 4.

⇒

⇒ m₁ ...(3)

Second curve is 4y + x² = 8

⇒ 4. + 2x = 0

⇒

⇒ m₂ ...(4)

Substituting (2,1) for m₁ & m_2,we get,

m₁

⇒

m₁ = 1 ...(5)

m₂

⇒

m₂ = – 1 ...(6)

when m₁ = 1 & m₂ = – 1

⇒ 1× – 1 = – 1

∴ Two curves x² = 4y & 4y + x² = 8 intersect orthogonally.

Given:

Curves x² = y ...(1)

& x³ + 6y = 7 ...(2)

The point of intersection of two curves (1,1)

Solving (1) & (2),we get,

First curve is x² = y

Differentiating above w.r.t x,

⇒ 2x

⇒

⇒ m₁ = 2x ...(3)

Second curve is x³ + 6y = 7

Differentiating above w.r.t x,

⇒ 3x² + 6. = 0

⇒

⇒

⇒ m₂ ...(4)

Substituting (1,1) for m₁ & m_2,we get,

m₁ = 2x

⇒ 2×1

m₁ = 2 ...(5)

m₂

⇒

m₂ = ...(6)

when m₁ = 2 & m₂ =

⇒ 2×1

∴ Two curves x² = y & x³ + 6y = 7 intersect orthogonally.

Given:

Curves y² = 8x ...(1)

& 2x² + y² = 10 ...(2)

The point of intersection of two curves are (0,0) & (1,2)

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ y² = 8x

⇒ 2y.8

⇒

⇒ ...(3)

⇒ 2x² + y² = 10

Differentiating above w.r.t x,

⇒ 4x + 2y. = 0

⇒ 2x + y. = 0

⇒ y. = – 2x

...(4)

Substituting (1,2)for m₁ & m_2,we get,

m₁

⇒

m₁ = ...(5)

m₂

⇒

m₂ = ...(6)

when m₁ & m₂

⇒ ×1

∴ Two curves y² = 8x & 2x² + y² = 10 intersect orthogonally.

Given:

Curves 4x = y² ...(1)

& 4xy = k ...(2)

We have to prove that two curves cut at right angles if k² = 512

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ 4x = y²

⇒ 4 = 2y.

⇒

m₁ ...(3)

⇒ 4xy = k

Differentiating above w.r.t x,

⇒ 4(1×) = 0

⇒ = 0

m₂ ...(4)

Since m₁ and m₂ cuts orthogonally,

⇒ ×1

⇒ 1

⇒ x = 2

Now , Solving (1) & (2),we get,

4xy = k & 4x = y²

⇒ (y²)y = k

⇒ y³ = k

⇒ y

Substituting y in 4x = y²,we get,

⇒ 4x = ()²

⇒ 4×2

⇒ 8

⇒ 8³

⇒ 512

Given:

Curves 2x = y² ...(1)

& 2xy = k ...(2)

We have to prove that two curves cut at right angles if k² = 8

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ 2x = y²

⇒ 2 = 2y.

⇒

m₁ ...(3)

⇒ 2xy = k

Differentiating above w.r.t x,

⇒ 2(1×) = 0

⇒ = 0

m₂ ...(4)

Since m₁ and m₂ cuts orthogonally,

⇒ ×1

⇒ 1

⇒ x = 1

Now , Solving (1) & (2),we get,

2xy = k & 2x = y²

⇒ (y²)y = k

⇒ y³ = k

⇒ y

Substituting y in 2x = y²,we get,

⇒ 2x = ()²

⇒ 2×1

⇒ 2

⇒ 2³

⇒ 8

Given:

Curves xy = 4 ...(1)

& x² + y² = 8 ...(2)

Solving (1) & (2),we get

⇒ xy = 4

⇒ x

Substituting x in x² + y² = 8,we get,

⇒ ()² + y² = 8

⇒ + y² = 8

⇒ 16 + y⁴ = 8y²

⇒ y⁴ – 8y² + 16 = 0

We will use factorization method to solve the above equation

⇒ y⁴ – 4y² – 4y² + 16 = 0

⇒ y²(y² – 4) – 4(y² – 4) = 0

⇒ (y² – 4)(y² – 4) = 0

⇒ y² – 4 = 0

⇒ y² = 4

⇒ y = ±2

Substituting y = ±2 in x,we get,

⇒ x

⇒ x = ±2

∴ The point of intersection of two curves (2,2) &

( – 2, – 2)

First curve xy = 4

⇒ 1×y + x. = 0

⇒ x. = – y

⇒ m₁ ...(3)

Second curve is x² + y² = 8

Differentiating above w.r.t x,

⇒ 2x + 2y. = 0

⇒ y. = – x

⇒ m₂ ...(4)

At (2,2),we have,

m₁

⇒

m₁ = – 1

At (2,2),we have,

⇒ m₂

⇒

⇒ m₂ = – 1

Clearly, m_{1 =} m₂ = – 1 at (2,2)

So, given curve touch each other at (2,2)

Given:

Curves y² = 4x ...(1)

& x² + y² – 6x + 1 = 0 ...(2)

∴The point of intersection of two curves is (1,2)

First curve is y² = 4x

Differentiating above w.r.t x,

⇒ 2y. = 4

⇒ y. = 2

⇒ m₁ ...(3)

Second curve is x² + y² – 6x + 1 = 0

⇒ 2x + 2y.6 – 0 = 0

⇒ x + y.3 = 0

⇒ y. = 3 – x

...(4)

At (1,2),we have,

m₁

⇒

m₁ = 1

At (1,2),we have,

⇒ m₂

⇒

⇒ m₂ = 1

Clearly, m_{1 =} m₂ = 1 at (1,2)

So, given curve touch each other at (1,2)

Given:

Curves 1 ...(1)

& xy = c² ...(2)

First curve is 1

Differentiating above w.r.t x,

⇒ . = 0

⇒ .

⇒

⇒

⇒ m₁ ...(3)

Second curve is xy = c²

⇒ ⇒ 1×y + x. = 0

⇒ x. = – y

⇒ m₂ ...(4)

When m₁ & m₂ =

Since ,two curves intersect orthogonally,

⇒ × = – 1

⇒ = 1

⇒ ∴ a² = b²

Given:

Curves 1 ...(1)

& 1 ...(2)

First curve is 1

Differentiating above w.r.t x,

⇒ . = 0

⇒ .

⇒

⇒

⇒ m₁ ...(3)