Conic Sections

the circle touches both the x and y axes in the first quadrant and the radius is a.

For a circle of radius a, the centre is (a,a).

The equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

Therefore, the equation of the circle becomes (x – a)² + (y – a)² = a²

x² - 2ax + a² + y² - 2ay + a² - a² = 0

x² - 2ax + y² - 2ay + a² = 0

Squaring both the equations,

Adding both the equations,

x²+y²= a²

The equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

Centre = (0, 0) Radius = a units

Hence proved.

The equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

Putting the values of given co-ordinates in the above expression,

(0,0)

(0 – h)² + (0 – k)² = r²

h² + k² = r²

(a,0)

(a – h)² + (0 – k)² = r²

a² + h² - 2ah + k² = r² ------ (1)

(0,b)

(0 – h)² + (b – k)² = r²

h² + b² + k² - 2bk = r² ------- (2)

On solving equations (1) & (2), respectively,

a (a – 2h) = 0

b (b – 2k) = 0

So,

a = 0 or 2h

b = 0 or 2k respectively.

Since the circle passes through the centre (0,0), so the co – ordinates are

a = 2h,

b = 2k,

Th co – ordinates of the centre are .

Since the circle has a centre (1,2) and also touches x-axis.

Radius of the circle is, r = 2

The equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

So, the equation of the required circle is:

(x – 1)² + (y – 2)² = 2²

⇒x² - 2x + 1 + y² - 4y + 4 = 4

⇒ x² + y² – 2x - 4y + 1 = 0

The equation of the circle is x² + y² – 2x - 4y + 1 = 0.

Since both lines are parallel & tangent to the circle. Then distance between these two lines must be the diameter of the circle.

Lines:

3x - 4y + 4 = 0

3x - 4y - 3.5 = 0 (Equating the co-efficient of the equation)

Distance (between 2 lines) = for the two given equations,

ax + by + c = 0 &ax + by + p = 0

Distance = = units

Radius =

=0.75 units.

Hence, the radius of the circle is 0.75 units.

As the circle lies in third quadrant, then the centre is (-a, -a).

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0

The line which touches the circle is 3x−4y+8=0, which is a tangent to the circle.

∴ The perpendicular distance = a units (radius of the circle)

Co-ordinates of the centre of the circle = (-2,-2)

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(x – (-2))² + (y – (-2))² = 2²

(x + 2)² + (y + 2)² = 4

x² + 4x + 4 + y² + 4y + 4 - 4 = 0

x² + y² + 4x + 4y + 4 = 0

The equation of the given circle is x² + y² + 4x + 4y + 4 = 0.

Equation of the circle,

x² - 4x + y² - 6y + 11 = 0

x² - 4x + 4 + y² - 6y + 9 +11 – 13 = 0

x² - 2(2)x + 2² + y² - 2(3)y + 3² +11 – 13 = 0

(x – 2)² + (y – 3)² = 2

(x – 2)² + (y – 3)² = (√2)²

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

Centre = (2,3)

the centre point is the mid-point of the two ends of the diameter of a circle.

Let the points be (p,q)

p + 3 = 4 & q + 4 = 6

p = 1 & q = 2

Hence, the other ends of the diameter are (1,2).

Solving the given equations,

3x + y = 14 ------- (1)

2x + 5y = 18 ------- (2)

Multiplying the first equation by 5,

15x + 5y = 70

2x + 5y = 18

13 x = 52, x = 4Putting x = 4, in first equation,

3(4) + y = 14

Y = 14 – 12 = 2

So, the point of intersection is (4,2)

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

Putting the values of (4,2) and centre co-ordinates (1,-2) in the above expression,

(4 – 1)² + (2 – (-2))² = r²

3² + 4² = r²

r² = 9 + 16 = 25

r = 5 units

So, the expression is

(x – 1)² + (y – (-2))² = 5²

x² - 2x + 1 + (y + 2)² = 25

x² - 2x + 1 + y² + 4y + 4 = 25

x² - 2x + y² + 4y – 20 = 0

Hence the required expression is x² - 2x + y² + 4y – 20 = 0.

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(x – 0)² + (y – 0)² = 4²

Perpendicular Distance between a point (0, 0) & the line or

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0

(Radius = 4, Given)

Hence, the required value of k is 8.

Equation of the circle,

x² - 6x + y² + 12y + 15 = 0

x² - 2(3)x + 3² + y² + 2(6)y + 6² + 15 – 9 + 36 = 0

(x – 3)² + (y-(-6))² - 30 = 0

(x – 3)² + (y-(-6))² = (√30)²

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

Centre = (3,-6)

Area of inner circle = units square

Area of outer circle = units square {Given}

So,

r² = 60

Equation of outer circle is,

(x – 3)² + (y-(-6))² = (√60)²

x² - 6x + 9 + y² - 12 y + 36 = 60

x² - 6x + y² +12y +45 – 60 = 0

x² - 6x + y² + 12y – 15 = 0

Hence, the required equation of the circle is x² - 6x + y² + 12y – 15 = 0.

Equation of an ellipse = , whereas

Length of latus rectum =

Length of minor axis = 2b

So,

{Given}

Ab = 2b²

2b² – ab = 0

b (2b - a) = 0

So, b = 0 or a = 2b

b² = a² (1 - e²)

b² = (2b)² (1 - e²)

b² = 4b² (1 - e²)

Hence, the eccentricity is .

Equation of an ellipse = , whereas

Length of latus rectum =

Length of minor axis = 2b

9x² + 25y² = 225

Dividing the equation by 225,

a = 5, b = 3

b² = a² (1 - e²)

3² = 5² (1 - e²)

Foci = (±ae,0)

=

Hence, the eccentricity is and foci is (±4,0).

Equation of an ellipse = , whereas

Length of latus rectum =

Length of minor axis = 2b

Foci = (±ae,0)

Distance between foci = 10 (Given)

2ae = 10

a = 8

b² = a² (1 - e²)

Length of Latus Rectum =

Hence, the length of latus rectum is 9.75 units.

Equation of an ellipse = , whereas

Length of latus rectum =

Length of minor axis = 2b

b² = a² (1 - e²) --------- (i)

Length of Latus Rectum =

b² = 2.5 a --------- (ii)

[Putting the values in equation (i)]

22.5 a = 5a²

5a² - 22.5a = 0

5a (a – 4.5) = 0

a = 0 {Not Possible, as the length of latus rectum is 5 units} or 4.5

[Putting the values in equation (ii)]

Equation of an ellipse =

Equation of an ellipse is

Equation of an ellipse = , whereas

Length of latus rectum =

Length of minor axis = 2b

b² = a² (1 - e²) --------- (i)

From equation (i),

20 = 36 (1 - e²)

Directrices =

Distance between directrices =

Hence the distance between directrices is 18 units.

Equation of an ellipse is y² = 4ax, whereas

Length of latus rectum = 4a

Comparing the equations,

4a = 8

a = 2

y²= 8×2 = 16

Y = ± 4 & x = 2

Hence the co - ordinates are (2,4) & (2, -4).

Equation of an ellipse is y² = 4ax, whereas

Length of latus rectum = 4a

Let the point on parabola be (i,j)

From the figure, slope of OP =

[Squaring the above mentioned equation] ------- (i)

j² = 4ai------- (ii)

Equating the equations, {(i) & (ii)}

i = 0 (not Possible) or

Now OP =

Hence, the length of line segment is units.

Vertex = (0,4) Focus = (0,2)

So, the directrix of the parabola is y = 6,

Since, Distance of (x, y) from (0, 2) and perpendicular distance from (x, y) to directrix are always equal.

Using Distance Formula & Perpendicular Distance Formula,

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0 i.e.., PM x(0) + y – 6 = 0 & (x, y)

Distance between the point of intersection & centre = [Distance Formula]

Squaring both the sides,

x² + y² - 4y + 4 = y² - 12y + 36

x² + 8y – 32 = 0

Hence, the required equation is x² + 8y – 32 = 0.

Solving the given equations,

y = mx + 1 & y² = 4x

(mx + 1)² = 4x

m²x² + 2mx + 1 = 4x

m²x² + 2mx – 4x + 1 = 0

m+x² + x (2m – 4) + 1 = 0

As the line touches the parabola, above equation must have equal roots,

Discriminant (D) = 0

(2m – 4)² - 4 (m²) (1) = 0

4m² - 16m + 16 – 4m² = 0

-16 m + 16 = 0

- m + 1 = 0

m = 1

Hence, the required value of m is 1.

Equation of Hyperbola =

Foci = (±ae, 0)

Distance between foci is 2ae = 16 (given)

e = √2

2 × a × √2 = 16

b² = a² (e² - 1)

Equation is

9y² - 4x² = 36

Dividing the equation by 36,

Equation of Hyperbola =

Comparing the equations,

a = 2 & b = 3

b² = a² (e² - 1)

3² = (2)²[(e)² - 1)]

9 = 4 (e² - 1)

Hence, the eccentricity of given hyperbola is .

Foci = (± ae,0) = (± 2, 0)

the hyperbola lies on x – axis,

Equation is

ae = 2 (given)

b² = a² (e² - 1)

Equation is

Hence, the required equation is

Since, diameters of a circle intersect at the centre of a circle,

2x – 3y = 5 -------- (i)

3x – 4y = 7 -------- (ii)

Solving the above mentioned equations,

6x – 9y = 15 [Multiplying equation (i) by 3}

6x – 8y = 14 [Multiplying equation (ii) by 2}

y = 1

y = -1

Putting y = -1, in equation (i)

2x – 3(-1) = 5

2x + 3 = 5

2x = 2

x = 1

Coordinates of centre = (1,-1)

Area = (Given)

r = 7 units

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(x – 1)² + (y – (-1))² = 7²

x² - 2x +1 + (y + 1)² = 49

x² - 2x + 1 + y² + 2y + 1 – 49 = 0

x² - 2x + y² + 2y – 47 = 0

Hence the required equation of the given circle is x² - 2x + y² + 2y – 47 = 0.

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²----------- (A)

Putting (2,3) & (4,5) in the above equation,

(2 – h)² + (3 – k)² = r²

4 – 4h + h² + 9 + k² - 6k = r²

h² - 4h + k² - 6k + 13 = r² --------- (i)

(4 – h)² + (5 – k)² = r²

16 – 8h + h² + 25 + k² - 10k = r²

h² - 8h + k² - 10k + 41 = r² --------- (ii)

Equating both the equations (i) & (ii), as their RHS are equal,

h² - 4h + k² - 6k + 13 = h² - 8h + k² - 10k + 41

8h - 4h + 10k – 6k = 41 – 13

4h + 4k = 28

h + k = 7 ------- (iii)

As centre lies on the given line, so it satisfies the values too,

k – 4h + 3 = 0 ---------- (iv)

Solving equations {(iii) & (iii)} simultaneously,

h + k = 7

-4h + k = -3

Subtracting both the equations,

(+) (-) (+)

5h = 10

h = 2

2 + k = 7

k = 5

Putting h = 2 & k = 5 in equation (i),

h² - 4h + k² - 6k + 13 = r²

2² - 4(2) + 5² - 6(5) + 13 = r²

4 – 8 + 25 – 30 + 13 = r²

r² = 4

r = 2 units

Putting the values of h = 2, k = 5 & r = 2, respectively in equation (A),

(x – h)² + (y – k)² = r²

(x – 2)² + (y – 5)² = 2²

x² - 4x + 4 + y² - 10y + 25 = 4

x² - 4x + y² - 10y + 25 =0

Hence, the required equation is x² - 4x + y² - 10y + 25 =0.

As the equation of the chord is,

2x – 5y + 18 = 0 ------- (i)

5y = 2x + 18

As, y = mx + C

Whereas, m is the slope of the line,

Slope of the line perpendicular to the chord,

As the product of slope of perpendicular lines = -1,

y - y₁ = m^' (x - x₁)

2y + 2 = -5x + 15

5x + 2y = 13 ------- (ii) [Equation of line passing from centre and cutting the chord]

Solving both the equations,

2x – 5y = -18 & 5x + 2y = 13

Multiplying the eq. (i) & eq. (ii) by 2 & 5 respectively,

4x – 10y = -36

25x + 10y = 65

29x = 29

x = 1

2(1) – 5y = -18

2 – 5y + 18 = 0

5y = 20

y = 4

Point of intersection @ chord and radius = (1,4)

Distance between the point of intersection & centre = [Distance Formula]

=√29 units

Using Pythagoras Theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

= (3)² + (√29)² = 29 + 9

= √38

Hypotenuse = √38 units (radius)

Since, the radius bisects the chord into two equal halves,

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(x – 3)² + (y –(-1))² = (√38)²

x² - 6x + 9 + (y + 1)² = 38

x² - 6x + y² + 2y + 1 + 9- 38 = 0

x² - 6x + y² + 2y – 28 = 0
Hence, the required equation of the circle is x² - 6x + y² + 2y – 28 = 0.

x² - 2x + y² - 4y – 20 = 0

x² - 2x + 1 +y² - 4y +4 – 20 – 5 = 0

(x – 1)² + (y – 2)² = 25

(x – 1)² + (y – 2)² = 5²

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

Centre = (1, 2)

Point of Intersection = (5, 5)

It intersects the line into 1: 1, as the radius of both the circles is 5 units.

Using Ratio Formula,

Ratio = m₁ : m₂

Assuming the co-ordinates of the centre of the circle be (p,q)

p + 1 = 10, q + 2 = 10

p = 9 & q = 8

Co-ordinates = (9,8)

Equation is,

(x – h)² + (y – k)² = r²

(x – 9)² + (y – 8)² = 5²

x² - 18x + 81 + y² - 16y + 64 = 25

x² - 18x + y² - 16y + 145 – 25 = 0

x² - 18x + y² - 16y + 120 = 0

Hence, the required equation is x² - 18x + y² - 16y + 120 = 0.

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²Centre lies on the line i.e., y = x – 1,

Co – Ordinates are (h, k) = (h, h – 1)

(x – h)² + (y – k)² = r²

(7 – h)² + (3 – (h – 1))² = 3²

49 + h² - 14h + (3 – h +1)² = 9

h² - 14h + 49 +16 +h² - 8h – 9 = 0

2h² - 22h + 56 = 0

h² - 11h + 28 = 0

h² - 4h – 7h + 28 = 0

h (h – 4) – 7 (h – 4) = 0

(h – 7) (h – 4) = 0

h = 7 or 4

Centre = (7, 6) or (4, 3)

(x – h)² + (y – k)² = r²

Equation, having centre (7, 6)

(x – 7)² + (y – 6)² = 3²

x² - 14x + 49 + y² - 12y + 36 – 9 = 0

x² - 14x + y² - 12y + 76 = 0

Equation, having centre (4, 3)

(x – 4)² + (y – 3)² = 3²

x² - 8x + 16 + y² - 6y + 9 – 9 = 0

x² - 8x + y² - 6y + 16 = 0

Hence, the required equation is x² - 14x + y² - 12y + 76 = 0 or x² - 8x + y² - 6y + 16 = 0.

the distance of any point on the parabola from its focus and its directrix is same.

Given that, directrix, x = 0 and focus = (6, 0)

If a parabola has a vertical axis, the standard form of the equation of the parabola is (x - h)² = 4p(y - k), where p≠ 0.

The vertex of this parabola is at (h, k).

The focus is at (h, k + p) & the directrix is the line y = k - p.

As the focus lies on x – axis,

Equation is y² = 4ax or y² = -4ax

So, for any point P(x, y) on the parabola

Distance of point from directrix = Distance of point from focus

x² = (x – 6)² + y²

x² = x² - 12x + 36 + y²

y² - 12x + 36 = 0

Hence the required equation is y² - 12x + 36 = 0.

Vertex = (0, 4) & Focus = (0, 2)

the distance between the vertex and directrix is same as the distance between the vertex and focus.

Directrix is y – 6 = 0

For any point of P(x, y) on the parabola

Distance of P from directrix = Distance of P from focus

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0 i.e.., x(0) + y – 6 = 0 & point = (x,y)

Distance between the point of intersection & centre = [Distance Formula] {Between (x,y) & (0,2)}

Squaring both the sides,

x² + y² - 4y + 4 = (y – 6)²

x² + y² - 4y + 4 = y² - 12y + 36

x² + 8y – 32 = 0
Hence, the required equation is x² + 8y – 32 = 0.

Focus = (–1, –2), directrix is x – 2y + 3 = 0

For any point (x,y) on parabola, the distance from focus to that point is always equal to the perpendicular distance from that point to the directrix,

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0 i.e.., x - 2y + 3 = 0 & point = (x,y)

Distance between the point of intersection & centre = [Distance Formula]

Squaring both the sides,

x² + 4y² + 9 - 4xy + 6x - 12y = 5 [x² + 2x + 1 + y² + 4y + 4]

x² + 4y² + 9 - 4xy + 6x- 12y = 5x² + 10x + 5y² + 20y + 25

4x² + y² + 4xy + 4x + 32y + 16 =0

Hence the required equation is 4x² + y² + 4xy + 4x + 32y + 16 = 0

It is given that the sum of the distances from the point (3,0) and (9,0) is 12.

Let the point be (x, y)

Distance between the point of intersection & centre = [Distance Formula]

Squaring both the sides,

Squaring both the sides,

x² + 12x + 36 = 4x² - 24x + 36 + 4y + 2

3x² - 36x + 4y²=0
Hence, the required equation is 3x² - 36x + 4y² = 0.

Let the point be P (x, y).

According to the question,

Distance of P from (x, y)(Distance from the line y=9)

Distance between the (x,y)& (0,4) = [Distance Formula]

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0 i.e.., x(0) + y – 6 = 0 & point = (x,y)

Distance between y – 9 = 0 {x(0) +y – 9 = 0} & (x, y)

Squaring both the sides,

x² + y² - 8y + 16 (y² - 18y + 81)

9x² + 9y² - 72y + 144 = 4y² - 72y + 324

9x² + 5y² = 180

Hence, the required equation is 9x² + 5y² = 180

, which is an ellipse.

Let the point be (x,y)

As per the question,

Distance of (x,y) from (-4,0) - Distance of (x,y) from (4, 0) = 2

Distance between the points = [Distance Formula]

Squaring both the sides,

Squaring both the sides,

16x² - 8x + 1 = x² - 8x + 16 + y²

15x² - y² = 15
Hence, the required equation is 15x ^� - y²=15, which is a parabola.

Vertices (± 5, 0), foci (± 7, 0)

a = 5 ae = 7

As, b²=a² (e²-1)

=24

General Equation of Hyperbola is

Hence, the general equation is .

Vertices (0, ± 7),

b = 7,

As, a² = b² (e² - 1)

=

General Equation of Hyperbola is

Hence, the required equation is .

Foci = & passing through (2, 3)

As

Also, a² = b² (e² - 1)

= b²e² - b²

=

= 10 - b²

General Equation of Hyperbola is

As, this hyperbola passed through (2, 3)

4b² - 9 (10 - b²) = -1(10 - b²) (b²)

4b² – 90 + 9b² = -10b² + b⁴

b⁴ - 23b² + 90 = 0

b⁴ - 18b² - 5b² + 90 = 0

b²(b² - 18)- 5 (b² - 18) = 0

(b² - 5) (b² - 18) =0

Either b² = 5 or b² = 18 (not possible, as a² + b² = 10)

a² + b² = 10

a² + 5 = 10

a² = 5

Hence, the required equation is .

x² + 6x + 9 + y² + 2y + 1 – 10 = 0

So, the centre points are (-3, -1)

Diameter Line has to pass through the mentioned centre points,

x + 3y = 0

(-3) + 3 (-1) = -6 ≠ 0

FALSE, as it (-3, -1) does not satisfy or lie on the diameter line.

x² - 14x + 49 + y² - 10y + 25 – 74 – 151 = 0

(x – 7)² + (y – 5)² = 225

(x – 7)² + (y – 5)² = 15²

Centre = (7, 5) Radius = 15 units

Putting x = 7 & y = 5 in the circle’s equation,

(7)² + 5² - 14 (7) – 10 (5) – 151

= -56

-56 < 0

(2, -7) lies inside the circle.

Distance from (2, -7) to (7, 5),

Distance between the point of intersection & centre = [Distance Formula]

D =

D =

D =

D =

Shortest Distance = Radius – Distance between centre & given point

= 15 – 13 = 2 units

Hence the answer is FALSE.

lx + my = 1

my = 1 – lx

Slope of tangent = -l

x² + y² = a²

(x – 0) ² + (y – 0) ² = a²

Centre = (0, 0) Radius = a units

Slope of line perpendicular to tangent= {Product of slopes of perpendicular lines = -1}

Equation of line perpendicular to tangent i.e.., Radius of the circle

ly = x – 0

Distance of centre (0,0) from the line lx + my = 1 is equal to the radius ‘a’.

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0

Squaring both the sides,

Hence, the locus of (l,m)is, which is a circle.

TRUE

Putting the values, in the given equation,

x² + y² - 2x + 6y + 1 = 0

1² + 2² - 2 (1) + 6 (2) + 1

1 + 4 – 2 + 12 + 1

= 16 > 0

So the point lies outside the circle.

FALSE

lx + my + n = 0, y² = 4ax & ln = am²

Solving both the above mentioned expressions,

Putting the value of in the equation of the line,

As the line touched parabola, above equations must have equal roots.

Discriminant (D) = 0

Hence proved.

TRUE

As the sum of distances of any point P on the ellipse from the two foci is equal to the length of the major axis.

Major Axis = 2b = 10

b = 5

S & S’ are foci, then PS + PS’ = 5 + 5 = 10 ≠ 8

FALSE

4x² + 9y² = 72 & 2x + 3y = 12

Solving both the expressions,

2x = 12 – 3y

Squaring both sides,

(2x)² = (12 – 3y)²

4x² = (12 – 3y)²

Putting the value of 4x²,

(12 – 3y)² + 9y² = 72

144 + 9y² - 72y + 9y² - 72 = 0

18y² - 72y + 72 = 0

Y² - 4y + 4 = 0

(y – 2)²= 0

y = 2

2x = 12 – 3y

2x = 12 – 3(2)

2x = 12 – 6 = 6

x = 3

So, the point of intersection is (3,2).

TRUE

-------- (i)

--------(ii)

Equating the values of k,

3x² - y² = 48

Dividing the equation by 48,

a² = 16 & b² = 48

As,

48 = 16 (e² - 1)

e² - 1 = 3

e² = 4

e = 2

Hence, the eccentricity of the given hyperbola is 2, which makes the statement correct, and it is TRUE.

Perpendicular Distance between a point (3, -4) & the line 5x +12y – 12 = 0,

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

169x² - 1014x + 1521 + 169y² + 1352y + 2704 – 2025 = 0

169x² - 1014x + 1521 + 169y² + 1352y + 679 = 0

Hence, the required equation is 169x²-1014x + 1521 + 169y² + 1352y + 679 = 0.

Solving the given equations,

y = x + 2 & 3y = 4x

3 (x + 2) = 4x

3x + 6 = 4x

x = 6, y = x + 2 = 6 + 2 = 8

Points of intersection is (6, 8)

y = x + 2 & 2y = 3x

2 (x + 2) = 3x

2x + 4 = 3x

x = 4, y = x + 2 = 4 + 2 = 6

Points of intersection is (4, 6)

3y = 4x & 2y = 3x

x = 0 y = 0

Points of intersection is (0, 0)

Expanded form of a circle,

x² + y² + 2gx + 2fy + C = 0, whereas Centre = (-g, -f)

(6, 8)

6² + 8² + 2g(6) + 2f(8) + C = 0

36 + 64 + 12g + 16f + C = 0

16f + 12g + C + 100 = 0

(4, 6)

4² + 6² + 2g(4) + 2f(6) + C = 0

16 + 36 + 8g + 12f + C = 0

12f + 8g + C + 52 = 0

(0, 0)

0² + 0² + 2g(0) + 2f(0) + C = 0

0 + 0 + 0 + 0 + C = 0

C + 0 = 0

C = 0

16f + 12g + 100 = 0

12f + 8g + 52 = 0

Solving the above mentioned equations, simultaneously,

48f + 36g + 300 = 0

48f + 32g + 208 = 0

4g + 92 = 0

g + 23 = 0

g = -23

16f + 12 (-23) + 100 = 0

16f – 276 + 100 = 0

16f = 176

f = 11 & g = -23

Expanded form of a circle,

x² + y² + 2gx + 2fy + C = 0, whereas Centre = (-g, -f)

x² + y² + 2(-23)x + 2(11)y + C = 0

x² + y² - 46x + 22y + C = 0

Hence, the required equation is x² + y² - 46x + 22y + C = 0.

a = 6, a² = 36

b = 4, b² = 16

b² = a² (1 - e²)

16 = 36 (1 - e²)

4 = 9 (1 - e²)

Hence the length of strings and distance between the pins is & respectively.

Foci = (0, ±ae) = (0, ±1)

be = 1

Minor Axis = 2a = 1, a = 0.5

a² = b² (1 - e² )

(0.5)² = b² (1 - e²)

0.25 = b² - b²e²

0.25 = b² - 1 {As be = 1}

Equation of ellipse is,

Hence, the required equation is .

S = Focus = (-1, -2) & Directrix is x – 2y + 3 = 0

Let a point on the parabola be P (x,y).

So, length of perpendicular on the directrix = Distance from point (x,y) to the Focus

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0

Using Distance Formula,

Squaring both the sides,

x² + 4y² + 9 - 4xy - 12y + 6x = 5x² + 10x + 25 + 5y² + 20y

4x² + 4x + 4xy + y² + 32y + 16 = 0
Hence the equation of parabola is 4x² + 4x+4xy+ y²+32y+16=0

Equation of Hyperbola be,

Vertices are (0, ± b) = (0, ± 6)

b= 6,

a² = b² (e² - 1)

Foci = (0, ± be) =
Hence the equation of hyperbola and its foci are & respectively.

Radius is the distance between the given co – ordinates,

Using Distance Formula,

.

R = 5 units

Area =

Option (C) is the answer.

When circle touches both the axes, the co – ordinates of the centre and its radius are equal in their magnitude,

h = k – r

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(3 – h)² + (6 – h)² = h²

9 + h² - 6h + 36 + h² - 12h = h²

h² - 18h + 45 = 0

h² - 15h – 3h + 45 = 0

h (h – 15) – 3 (h – 15) = 0

(h – 3) (h – 15) = 0

h = 3 or h = 15

Co – ordinates of centre are (3, 3) or (15, 15)

(x – h)² + (y – k)² = r²

Equation, having centre (3, 3)

(x – 3)² + (y – 3)² = 3²

x² - 6x + 9 + y² - 6y + 9 – 9 = 0

x² - 6x + y² - 6y + 9 = 0

Equation, having centre (15, 15)

(x – 15)² + (y – 15)² = 15²

x² - 30x + 225 + y² - 30y + 225 – 225 = 0

x² - 30x + y² - 30y + 225 = 0

Hence the equations are x² - 6x + y² - 6y + 9 = 0 or x² - 30x + y² - 30y + 225 = 0.

Option (C) is the answer.

As the centre lies on y – axis, then the centre is (0, k)

radius = ‘k’ units

Using Distance Formula, {Between (2, 3) & (0, k)}

Equating the equations,
r = k &

k =

k² = k² - 6k + 13

6k = 13

Centre =

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

3x² + 3y² - 13y = 0

In triangle LMD,

(LM)² = (LD)² + (MD)² - Pythagoras Theorem,

LD = 3a (Given) &

(LM)² = 12a²

In triangle OMD,

(OM)² = (OD)² + (MD)² - Pythagoras Theorem,

As LO + OD = LD = 3a,

[Putting (LM)² = 12a²]

9a² + 3a² - 6aR = 0

6aR = 12a²

R = 2a

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(x – 0)² + (y – 0)² = (2a)²

x² + y² = 4a²

Hence the required equation is x² + y² = 4a ^�.

Option (D) is the answer.

As focus lies on y – axis,

Equation is x² = 4ay or x² = -4ay,

Focus has a –ve co = ordinate, so equation is x² = -4ay,

Co – Ordinate of focus = (0, -a)

-a = -3

a = 3

As per the equation,

x² = -4ay

x² = -4 (3) y

x² = -12y

Hence, the required equation is x² = -12y.

Option (A) is the answer.

y² = 4ax

2² = 4a (3)

4 = 12a

Length of Latus Rectum = 4a =

Hence, the length of latus rectum is units.

Vertex = (-3, 0) & Directrix is x + 5 = 0 (Intersection point is (-5, 0)

As , the co-ordinates of focus is the mid – point of the co-ordinates of vertex & the point of intersection of directrix and x or y axis respectively.

Focus = (p,q)

p = -1 & q = 0

Focus is (-1, 0)

For any point on parabola, P(x,y) the distance of focus from that point is always equal to the perpendicular distance from that point to the directrix.

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0 x + 5 = 0

Using Distance Formula,

Squaring both the sides,

x² + 2x + 1 + y² = x² + 25 + 10x

y² = 8x + 24

y² = 8 (x + 3)

Hence the required equation is y² = 8 (x + 3).

OPTION (A) is the correct answer.

Focus (S) = (1, -1) Directrix is x – y – 3 = 0 &

As, for any point P (x,y) on the ellipse,

Distance from focus to the point (x,y) = e (Distance from (x,y) to the foot of perpendicular on directrix)

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0

Using Distance Formula,

Squaring both the sides,

8[x² - 2x + 1 + y² + 2y + 1] = x² - 6x - 2xy + y² + 6y + 9

8x² - 16x + 16 + 8y² + 16y = x² - 6x - 2xy + y² + 6y + 9

7x² - 10x + 2xy + 7y² + 10y + 7 = 0

Hence the required equation is 7x² - 10x + 2xy + 7y² + 10y + 7 = 0

OPTION (A) is the correct answer.

3x² + y² = 12,

Dividing the equation by 12,

General Equation of Ellipse :

a = 2 &

Length of Latus Rectum =

Hence, the length of latus rectum is 12 units.

None of the options given, are correct.

b² = a² (1 - e²)

Hence, Option A is correct.

Length of Latus Rectum =

b² = 4a ------- (i)

As, Conjugate axis is equal to half of the distance between the foci,

2b = ae

Squaring the above mentioned equation,

4b² = a²e² -------- (ii)

b² = a² (1 - e²)

4a = a² (1 - e²)

4 = a (1 - e²)

-------- (iii)

Putting the value from equation (ii),

a = 20

b² = 4a = 4 × 20 = 80 {From equation (i)}

Hence, the eccentricity of hyperbola is .

Equation of Hyperbola is

Foci = (±ae, 0)

Distance between foci = 2ae = 16 (Given)

b² = a² (e² - 1)

Hence equation is .

Foci = (±ae, 0) = (±2, 0)

So, after comparing the equations,

ae = 2

b² = a² (e² - 1)

Equation of Hyperbola is

Hence is the required equation.