Conic Sections

The equation of a circle with centre (h,k)

and radius r is given as (x-h)²+(y-k)²=r²

It is given that centre (h,k)=(0, 2) and radius (r)=2.

Therefore, the equation of the circle is

(x-0)²+(y-2)²=2²

⇒ x²+y²+4-4y=4

⇒ x²+y² -4y=0

Therefore, the equation of the circle is x²+y² -4y=0

The equation of a circle with centre (h,k) and radius r is given as

(x-h)² + (y-k)²=r²

It is given that centre (h,K) = (-2,3) and radius (r) =4

Therefore, the equation of the circle is

(x+2)² + (y-3)² = (4)²

⇒ x² + 4x + 4 + y² - 6y + 9 = 16

⇒ x² + y² + 4x - 6y - 3 = 0

Therefore, the equation of the circle is x² + y² + 4x - 6y - 3 = 0

The equation of a circle with centre (h,k) and radius r is given as

(x-h)² + (y-k)²=r²

It is given that centre (h,k)= and radius(r) = .

Therefore, the equation of the circle is

⇒ x² – x + + y² -

⇒ x² – x + + y² -

⇒ x² – x + + y² -=0

⇒ 144x² -144x + 36 + 144y² -72y + 9- 1=0

⇒ 144x² -144x+ 144y² – 72y + 44 =0

⇒ 36x² +36x+ 36y² -18y+11 = 0

⇒ 36x² +36y² -36x – 18y+ 11= 0

The equation of a circle with centre (h,k) and radius r is given as

(x-h)² + (y-k)²=r²

It is given that centre (h,k) =(1, 1) and radius (r) =.

Therefore, the equation of the circle is

(x-1)² + (y-1)²=

⇒ x² – 2x + 1 + y² -2y + 1 =2

⇒ ² + y² -2x -2y=0

The equation of a circle with centre (h,k) and radius r is given as

(x-h)² + (y-k)²=r²

It is given that centre (h,k) = (-a, -b) and radius (r) =

(x + a)² +(y + b)² =

⇒ x² + 2ax + a² + y² + 2by + b² = a² – b²

⇒ x² + y² +2ax + 2by + 2b² = 0

The equation of the given circle is (x+5)² +(y-3)² = 36.

(x+5)² +(y-3)² = 36

+ (y - 3)² =6², which is of the form (x – h)² + (y – k )² = r²,

Where, h =-5, k =3 and r= 6.

Thus, the centre of the given circle is (-5, 3), while its radius is 6.

The equation of the given circle is x² + y² – 4x – 8y – 45 = 0.

x² + y² – 4x – 8y – 45 = 0

(x² – 4x) + (y² -8y) =45

+ – 4 – 16 = 45

(x - 2)² + (y – 4)² =65

(x - 2)² + (y – 4)² =, which is form (x-h)² +(y-k)² = r², where h=2,

K= 4 and r=

Thus, the centre of the given circle is (2, 4), while its radius is.

The equation of the given circle is x² + y² -8x + 10y -12 =0.

x² + y² -8x + 10y -12 =0

(x² – 8x) + (y²+10y) = 12

+ – 16 – 25 = 12

(x - 4)² + (y + 5)² =53

(x - 4)² + =, which is form (x-h)² +(y-k)² = r², where h=4,

K= -5 and r=

The equation of the given of the circle 2x² + 2y² –x =0.

2x² + 2y² –x =0

(2x² + x) +2y² =0

+ y² - = 0

+ (y - 0)² = , which is of the form (x-h)² + (y-k)² =r²,

Where, h = , k = 0, and r=

Thus, the centre of the given circle is , while its radius is .

Let the equation of the required circle be (x – h)²+ (y – k)² =r²

Since, the circle passes through points (4,1) and (6,5),

(4 – h)²+ (1 – k)² =r² .................(1)

(6– h)²+ (5 – k)² =r² ..................(2)

Since, the centre (h,k) of the circle lies on line 4x+y = 16,

4h + k =16..................... (3)

From the equation (1) and (2), we obtain

(4 – h)²+ (1 – k)² =(6 – h)² + (5 – k)²

16 – 8h + h² +1 -2k +k² = 36 -12h +h²+15 – 10k + k²

16 – 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k =11................ (4)

On solving equations (3) and (4), we obtain h=3 and k= 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)²+ (1 – 4)² =r²

(1)² + (-3)² = r²

1+9 = r²

r =

Thus, the equation of the requires circle is

(x – 3)²+ (y – 4)² =

⇒ x² -6x + 9 + y² – 8y + 16 =10

⇒ x² + y² -6x – 8y = 15 =0

Let the equation of the required circle be (x – h)²+ (y – k)² =r²

Since, the circle passes through points (2,3) and (-1,1).

(2 – h)²+ (3 – k)² =r² .................(1)

(-1 – h)²+ (1– k)² =r² ..................(2)

Since, the centre (h,k) of the circle lies on line x - 3y - 11= 0,

h - 3k =11..................... (3)

From the equation (1) and (2), we obtain

(2 – h)²+ (3 – k)² =(-1 – h)² + (1 – k)²

⇒ 4 – 4h + h² +9 -6k +k² = 1 + 2h +h²+1 – 2k + k²

⇒ 4 – 4h +9 -6k = 1 + 2h + 1 -2k

⇒ 6h + 4k =11................ (4)

Now multiply (3) by 6 and subtract it from (4) to get,

6h+ 4k - 6(h-3k) = 11 – 66

⇒ 6h + 4k – 6h + 18k = 11 – 66

⇒ 22 k = - 55

⇒ K = -5/2

Put this value in (4) to get,

6h + 4(-5/2) = 11

⇒ 6h – 10 = 11

⇒ 6h = 21

⇒ h = 21/6

⇒ h = 7/2

Thus we obtain h= and k= .

On substituting the values of h and k in equation (1), we get

+= r²

+= r²

+ = r²

+ = r²

Thus, the equation of the required circle is

+=

+=

⇒ 4x² -28x + 49 +4y² + 20y + 25 =130

⇒ 4x² +4y² -28x + 20y - 56 =0

⇒ 4(x² +y² -7x + 5y – 14) = 0

⇒ x²+y²-7x + 5y– 14 =0

Let the equation of the required circle be (x – h)²+ (y – k)² =r²

Since the radius of the circle is 5 and its centre lies on the x-axis, k= 0 and r= 5.

Now, the equation of the circle becomes (x – h)² + y² = 25.

It is given that the circle passes through the point (2, 3) so the point will satisfy the equation of the circle.

(2 – h)²+ 3² = 2

(2 – h)² = 25-9

(2 – h)² = 16

2 –h == 4

If 2-h =4, then h = -2

If 2-h = -4, then h =6.

When h= -2, the equation of the circle becomes

(x – 2)² + y² = 25

⇒ x² -12x + 36 + y² = 25

⇒ x² + y² -4x + 21 =0

When h= 6, the equation of the circle becomes

(x – 6)² + y² = 25

⇒ x² -12x + 36 + y² = 25

⇒ x² + y² -12x + 11 =0

Let the equation of the required circle be (x – h)²+ (y – k)² =r²

Since, the circle passes through (0, 0),

(0 – h)²+ (0 – k)² =r²

h² + k² = r²

The equation of the circle now becomes (x – h)²+ (y – k)² = h² + k².

It is given that the circle makes intercepts a and b on the coordinate axes.

That means that the circle passes through points (a, 0) and (0,b). Therefore,

(a – h)²+ (0 – k)² =h² +k².................(1)

(0 – h)²+ (b– k)² =h² +k²..................(2)

From equation (1), we obtain

a² – 2ah + h² +k² = h² +k²

a² – 2ah = 0

a(a – 2h) =0

a = 0 or (a -2h) = 0

However, a 0; hence, (a -2h) = 0

h =.

From equation (2), we obtain

h² – 2bk + k² + b²= h² +k²

b² – 2bk = 0

b(b– 2k) = 0

b= 0 or (b-2k) =0

However, a 0; hence, (b -2k) = 0

k =.

+=

⇒ 4x² -4ax + a² +4y² - 4by + b² =a² + b²

⇒ 4x² + 4y² -4ax – 4by =0

⇒ 4( x² +y² -7x + 5y – 14 ) = 0

⇒ x²+y²- ax - by=0

The centre of the circle is given as (h,k) = (2,2).

Since, the circle passes through point (4,5), the radius (r) of the circle is the distance between the points (2,2) and (4,5).

r =


=


=

=

Thus, the equation of the circle is

(x– h)²+ (y – k)² =r²

(x –h)² + (y – k)² =


(x –2)² + (y – 2)² =

x² - 4x+4 +y²- 4y₊4 =13

x² +y² – 4x -4y = 5

The equation of the given circle is x² +y² = 25.

x² +y² = 25

(x– 0)² +(y – 0)² = 5², which is of the form (x – h)²+ (y – k)² =r²,

Where, h=0, k=0 and r=5.

Distance between point (-2.5, 3.5) and the centre (0,0)

=

=

=

=4.3 (approx.) <5

Since, the distance between point (-2.5, -3.5) and the centre (0, 0) of the circle is less than the radius of the circle, point (-2.5, -3.5) lies inside the circle.

The given equation is y² = 12x

Here, the coefficient of x is positive.

Hence, the parabola opens towards the right.

On comparing this equation with y² = 4ax, we get,

4a = 12

⇒ a = 3

Thus,

Co-ordinates of the focus = (a, 0) = (3, 0)

Since, the given equation involves y², the axis of the parabola is the x-axis.

Equation of directrix, x =-a, then,

x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12

The given equation is y² = 6y

Here, the coefficient of y is positive.

Hence, the parabola opens upwards.

On comparing this equation with y² = 4ay, we get,

4a = 6

⇒ a =

Thus,

Co-ordinates of the focus = (0,a) =

Since, the given equation involves x², the axis of the parabola is the y-axis.

Equation of directrix, y =-a, then,

y =

Length of latus rectum = 4a = 6

The given equation is y² = -8x

Here, the coefficient of x is negative.

Hence, the parabola open towards the left.

On comparing this equation with y² = -4ax, we get,

-4a = -8

⇒ a = 2

Thus,

Co-ordinates of the focus = (-a,0) = (-2, 0)

Since, the given equation involves y², the axis of the parabola is the x-axis.

Equation of directrix, x =a, then,

y = 2

Length of latus rectum = 4a = 8

The given equation is x² = -16y

Here, the coefficient of y is negative.

Hence, the parabola opens downwards.

On comparing this equation with x² = -4ay, we get,

-4a = -16

⇒ a = 4

Thus,

Co-ordinates of the focus = (0,-a) = (0,-4)

Since, the given equation involves x², the axis of the parabola is the y-axis.

Equation of directrix, x =a, then,

y = 4

Length of latus rectum = 4a = 16

The given equation is y² = 10x

Here, the coefficient of x is positive.

Hence, the parabola open towards the right.

On comparing this equation with y² = -4ax, we get,

4a = 10

⇒ a =

Thus,

Co-ordinates of the focus = (a,0) =

Since, the given equation involves y², the axis of the parabola is the x-axis.

Equation of directrix, x =-a, then,

x =

Length of latus rectum = 4a = 10

The given equation is x² = -9y

Here, the coefficient of y is negative.

Hence, the parabola open downwards.

On comparing this equation with x² = -4ay, we get,

-4a = -9

⇒ a =

Thus,

Co-ordinates of the focus = (0,-a) =

Since, the given equation involves x², the axis of the parabola is the y-axis.

Equation of directrix, y = a, then,

x =

Length of latus rectum = 4a = 9

Focus (6,0); directrix x = -6

Since, the focus lies on the x–axis is the axis of the parabola.

Thus,

The equation of the parabola is either of the form y² = 4ax or y² = -4ax.

It is also seen that the directrix, x = -6 is to the left of the y- axis, while the focus (6, 0) is to the right of the y –axis.

Hence, the parabola is of the form y² = 4ax.

Here, a = 6

Thus, the equation of the parabola is y² = 24x.

Focus (0, -3); directrix y = 3

Since, the focus lies on the y–axis, the y-axis is the axis of the parabola.

Thus,

The equation of the parabola is either of the form x² = 4ay or x² = -4ay.

It is also seen that the directrix, y = 3 is above the x- axis, while the focus (0,-3) is below the x-axis.

Hence, the parabola is of the form x² = -4ay.

Here, a = 3

Thus, the equation of the parabola is x² = -12y.

Vertex (0, 0); focus (3, 0)

Since, the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y²=4ax.

Since, the focus is (3, 0), a = 3

Thus, the equation of the parabola is y² = 4 × 3 × x,

=> y² = 12x

Vertex (0, 0); focus (-2, 0)

Since, the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y²=-4ax.

Since, the focus is (-2, 0), a = 2

Thus, the equation of the parabola is y² = -4 × 2 × x,

=> y² = -8x

Since, the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the from y²=4ax or y² = -4ax.

The parabola passes through point (2, 3), which lies in the first quadrant.

Thus, the equation of the parabola is of the form y² = 4ax, while point (2, 3) must satisfy the equation y² = 4ax.

Thus,

3² = 4a(2)

⇒ a = 9/8

Thus, the equation of the parabola is

⇒ 2y² = 9x

Since, the vertex is (0, 0) and the parabola is symmetric about the y-axis, the equation of the parabola is either of the from x²=4ay or x² = -4ay.

The parabola passes through point (5, 2), which lies in the first quadrant.

Thus, the equation of the parabola is of the form x² = 4ay, while point (5, 2) must satisfy the equation x² = 4ay.

Thus,

5² = 4a(2)

⇒ 25 = 8a

⇒ a = 25/8

Thus, the equation of the parabola is

⇒ 2x² = 25y

The given equation is

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with =1, we obtain a =6 and b =4.

Therefore,

The coordinates of the foci are (2, 0) and (-6, 0).

The coordinates of the vertices are (6, 0) and (-6, 0)

Length of major axis = 2a =12

Length of minor axis = 2b =8

Eccentricity, e

Length of latus rectum =

The given equation is or

Here, the denominator of is greater than the denominator of.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with , we obtain b =2 and a =5.

c = =

Therefore,

The coordinates of the foci are (0, ) and( 0, - ) .

The coordinate of the vertices are (0, 5) and (0, -5)

Length of major axis = 2a =10

Length of minor axis = 2b =4

Eccentricity, e

Length of latus rectum =

The given equation is =1 or =1.

Here, the denominator of is greater than the denominator of.

Therefore, the major axis is along the x-axis , while the minor axis is along the y-axis.

On comparing the given equation with =1, we obtain b =4 and a =3.

c = =

Therefore,

The coordinates of the foci are ().

The coordinate of the vertices are ().

Length of major axis = 2a =8

Length of minor axis = 2b =6

Eccentricity, e

Length of latus rectum =

The given equation is =1 or =1.

Here, the denominator of is greater than the denominator of.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with =1, we obtain b =5 and a =10.

c = = = 5

Therefore,

The coordinates of the foci are ().

The coordinate of the vertices are (0,).

Length of major axis = 2a = 20

Length of minor axis = 2b = 10

Eccentricity, e

Length of latus rectum =

The given equation is =1 or =1.

Here, the denominator of is greater than the denominator of

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with =1, we obtain a =7 and b =6.

c = =

Therefore,

The coordinates of the foci are ().

The coordinate of the vertices are ( ).

Length of major axis = 2a = 14

Length of minor axis = 2b = 12

Eccentricity, e =

Length of latus rectum =

The given equation is =1 or =1.

Here, the denominator of is greater than the denominator of.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with =1, we obtain b =10 and a =20.

c = = = 10

Therefore,

The coordinates of the foci are ().

The coordinate of the vertices are (0,).

Length of major axis = 2a = 40

Length of minor axis = 2b = 20

Eccentricity, e

Length of latus rectum =

The given equation is 36x² + 4y² = 144.

It can be written as

36x² + 4y² = 144

Or, =1

Or, =1. …………(1)

Here, the denominator of is greater than the denominator of.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with =1, we obtain b =2 and a =6.

c = = = 4

Therefore,

The coordinates of the foci are ().

The coordinate of the vertices are (0, ).

Length of major axis = 2a = 12

Length of minor axis = 2b = 4

Eccentricity, e

Length of latus rectum =

The given equation is 16x² + y² = 16.

It can be written as

16x² + y² = 16

Or, =1

Or, =1. ………… (1)

Here, the denominator of is greater than the denominator of.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with =1, we obtain b =1 and a =4.

c = =

Therefore,

The coordinates of the foci are ().

The coordinate of the vertices are (0, ).

Length of major axis = 2a = 8

Length of minor axis = 2b = 2

Eccentricity, e

Length of latus rectum =

The given equation is 4x² + 9y² = 36.

It can be written as

4x² + 9y² = 36

Or, =1

Or, =1. ………… (1)

Here, the denominator of is greater than the denominator of.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with =1, we obtain a =3 and b =2.

c = =

Therefore,

The coordinates of the foci are ().

The coordinate of the vertices are ( ).

Length of major axis = 2a = 6

Length of minor axis = 2b = 4

Eccentricity, e

Length of latus rectum =

Vertices (), foci ( ).

Here, the vertices are on the x-axis.a

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, a=5 and c = 4.

It is known that a² = b²+c².

5² = b²+4²

25 = b² + 16

b² = 25 – 16

b = = 3

Thus, the equation of the ellipse is =1 or = 1.

Vertices (), foci (0,).

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, a=13 and c = 5.

It is known that a² = b²+c².

13² = b²+5²

169 = b² + 15

b² = 169 – 125

b = = 12

Thus, the equation of the ellipse is =1 or = 1.

Vertices (6, 0), foci (4, 0).

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, a=6 and c = 4.

It is known that a² = b²+c².

6² = b²+4²

36 = b² + 16

b² = 36 – 16

b =

Thus, the equation of the ellipse is =1 or = 1.

Ends of major axis (), ends of minor axis ().

Here, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, a=3 and b = 2.

Thus, the equation for the ellipse = 1 i.e., = 1.

Ends of major axis (0,), ends of minor axis ().

Here, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, a= and b = 1.

Thus, the equation for the ellipse = 1 or, = 1.

Length of major axis = 26; foci = ().

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, 2a =26 a = 13 and c=5.

It is known that a² = b²+c².

13² = b²+4²

169 = b² + 25

b² = 169 – 25

b = = 12

Thus, the equation of the ellipse is =1 or = 1.

Length of major axis = 26; foci = ().

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, 2b =16 b = 8 and c=6.

It is known that a² = b²+c².

a² = 8²+6^{2 =} 64+36 =100

b = = 10

Thus, the equation of the ellipse is =1 or = 1.

Foci = (), a=4

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, c = 3 and a=4.

It is known that a² = b²+c².

a² = 8²+6^{2 =} 64+36 =100

16 = b² + 9

b² = 16 – 9 = 7

Thus, the equation of the ellipse is = 1.

It is given that b = 3, c=4, centre at the origin; foci on the x axis.

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form =1, where a is the semi-major axis.

Accordingly, b = 3 and c=4.

It is known that a² = b²+c².

a² = 3²+4² = 9+16 =25

a =5

Thus, the equation of the ellipse is or = 1.

Since the center is at (0,0) and the major axis is on the y- axis, the equation of the ellipse will be of the form

=1 ………………….(1)

Where, a is the semi - major axis

The ellipse passes through points (3, 2) and (1, 6). Hence,

=1 …………………… (2)

=1……………………… (3)

On solving equation (2) and (3), we obtain b² = 10 and a² = 40.

Thus, the equation of the ellipse is = 1 or 4x² + y ² = 40.

Since the major axis is on the x-axis, the equation of the ellipse will be the form

=1 …………………. (1)

Where, a is the semi - major axis

The ellipse passes through points (4, 3) and (6,2). Hence,

=1 …………………… (2)

……………………… (3)

On solving equation (2) and (3), we obtain a² = 52 and b² = 13.

Thus, the equation of the ellipse is = 1 or x² + 4y² = 52.

The given equation is or

On comparing this equation with the standard equation of hyperbola

, we get,

a = 4 and b = 3,

We know, a² + b² = c²

Thus,

c² = 4² + 3² =25

⇒ c = 5

Therefore,

The coordinates of the foci are (5,0).

The coordinates of the vertices are (4, 0).

Eccentricity, e =

Length of latus rectum =

The given equation is or

On comparing this equation with the standard equation of hyperbola

, we get,

a = 4 and b =,

We know, a² + b² = c²

Thus,

c² = 3² + = 36

⇒ c = 6

Therefore,

The coordinates of the foci are (0, 6).

The coordinates of the vertices are (0, 3).

Eccentricity, e =

Length of latus rectum =

The given equation is 9y² – 4x² = 36

We can re-write the given as

or ………..(1)

On comparing this equation (1) with the standard equation of hyperbola

, we get,

a = 2 and b =3

We know, a² + b² = c²

Thus,

c² = 4 + 9 = 13

⇒ c =

Therefore,

The coordinates of the foci are (0, ).

The coordinates of the vertices are (0, 2).

Eccentricity, e =

Length of latus rectum =

The given equation is 16y² – 9x² = 576

We can re-write the given as

Or ……….. (1)

On comparing this equation (1) with the standard equation of hyperbola

, we get,

a = 6 and b =8

We know, a² + b² = c²

Thus,

c² = 36 + 64 = 100

⇒ c = 10

Therefore,

The coordinates of the foci are (10, 0).

The coordinates of the vertices are (6, 0).

Eccentricity, e =

Length of latus rectum =

The given equation is 5y² – 9x² = 36

We can re-write the given as

Or ……….. (1)

On comparing this equation (1) with the standard equation of hyperbola

, we get,

a = and b =2

We know, a² + b² = c²

Thus,

c² = =

⇒ c =

Therefore,

The coordinates of the foci are

The coordinates of the vertices are .

Eccentricity, e =

Length of latus rectum =

The given equation is 49y² – 16x² = 784

We can re-write the given as

Or ……….. (1)

On comparing this equation (1) with the standard equation of hyperbola

, we get,

a = 4 and b =7

We know, a² + b² = c²

Thus,

c² = 16 + 49 = 65

⇒ c =

Therefore,

The coordinates of the foci are (0, ).

The coordinates of the vertices are (0, 4).

Eccentricity, e =

Length of latus rectum =

Vertices ( 2, 0), foci ( 3, 0)

Here, the vertices are on the x-axis.

Thus,

The equation of the hyperbola is of the form

Since, the vertices are (2,0), a = 2

Since, the foci are (3, 0), c= 3

We know that, a² + b² = c²

Thus, 2² + b² = 3²

⇒ b² = 9 – 4 = 5

Hence, the equation of the hyperbola is

Vertices (0, 5), foci (0, 8)

Here, the vertices are on the y-axis.

Thus,

The equation of the hyperbola is of the form

Since, the vertices are (0, 5), a = 5

Since, the foci are (0, 8), c= 8

We know that, a² + b² = c²

Thus, 5² + b² = 8²

⇒ b² = 64 – 25 = 39

Hence, the equation of the hyperbola is

Vertices (0, 3), foci (0, 5)

Here, the vertices are on the y-axis.

Thus,

The equation of the hyperbola is of the form

Since, the vertices are (0, 3), a = 3

Since, the foci are (0, 5), c= 5

We know that, a² + b² = c²

Thus, 3² + b² = 5²

⇒ b² = 25 – 9 = 16

Hence, the equation of the hyperbola is

Foci ( 5, 0), the transverse axis is of length 8.

Here, the foci are on x-axis.

Thus,

The equation of the hyperbola is of the form

Since, the foci are (5, 0), c= 5

Since, the length of the transverse axis is 8,

2a = 8 ⇒ a = 4

We know that, a² + b² = c²

4² + b² = 5²

⇒ b² = 25 – 16 = 9

Hence, the equation of the hyperbola is

Foci (0, 13), the conjugate axis is of length 24.

Here, the foci are on y-axis.

Thus,

The equation of the hyperbola is of the form

Since, the foci are (0, 13), c= 13

Since, the length of the conjugate axis is 24,

2b = 24 ⇒ b = 12

We know that, a² + b² = c²

a² + 12² = 13²

⇒ a² = 169 – 144 = 25

Hence, the equation of the hyperbola is

Foci ( 3, 0), the latus rectum is of length 8.

Here, the foci are on x-axis.

Thus,

The equation of the hyperbola is of the form

Since, the foci are (), c=

Length of latus rectum = 8

⇒

⇒ b² = 4a

We know that, a² + b² = c²

a² + 4a = 45

⇒ a² + 4a -45 = 0

⇒ a² + 9a – 5a -45 = 0

⇒ (a +9)(a -5) = 0

⇒ A = -9,5

Since, a is non – negative, a = 5

Thus, b² = 4a = 4 × 5 = 20

Hence, the equation of the hyperbola is

Foci ( 4, 0), the latus rectum is of length 12

Here, the foci are on x-axis.

Thus,

The equation of the hyperbola is of the form

Since, the foci are ( 4, 0), c = 4

Length of latus rectum = 12

⇒

⇒ b² = 6a

We know that, a² + b² = c²

a² + 6a = 16

⇒ a² + 6a -16 = 0

⇒ a² + 8a – 2a - 16 = 0

⇒ (a +8)(a -2) = 0

⇒ A = -8, 2

Since, a is non – negative, a = 2

Thus, b² = 6a = 6 × 2 = 12

Hence, the equation of the hyperbola is

vertices ( 7,0), e =

Here, the vertices are on the x- axis

Thus, the equation of the hyperbola is of the form

Since, the vertices are ( 7,0), a = 7

It is given that e =

We know that, a² + b² = c²

Thus,

7² + b² =

Thus, the equation of the hyperbola is

Foci (0, ), passing through (2, 3)

Here, the foci are on y-axis.

Thus,

The equation of the hyperbola is of the form

Since, the foci are (, 0), c =

We know that, a² + b² = c²

⇒ b² = 10 – a² …………..(1)

Since, the hyperbola passes through point (2, 3)

……………(2)

From equations (1) and (2), we get,

⇒ 9(10 – a²) – 4a² = a²(10 –a²)

⇒ 90 – 9a² – 4a² = 10a² – a⁴

⇒ a⁴ -23a² + 90 = 0

⇒ a⁴ -18a² -5a²+ 90 = 0

⇒ a²(a² -18) -5(a² -18) = 0

⇒ (a² – 18)(a² -5) = 0

⇒ a² = 18 or 5

In hyperbola, c>a that is c²> a²

Thus, a² = 5

⇒ b² = 10 – a² = 10 – 5 = 5

Hence, the equation of the hyperbola is

The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x – axis.

This can be diagrammatically represented as

The equation of the parabola is of the form y² = 4ax (as it is opening to the right)

Since, the parabola passes through point A(10, 5), 10² = 4a(5)

⇒ 100 = 20a

⇒ a = = 5

Therefore, the focus of the parabola is (a, 0) = (5,0), which is the mid – point of the diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the negative y –axis.

This can be diagrammatically represented as

The equation of the parabola is of the form x² = -4ay (as it is opening downwards).

It can be clearly seen that the parabola passes through point

Therefore, the arch is in the form of a parabola whose equation is

When, y = -2,

Hence, when the arch is 2m from the vertex of the parabola, its width is approximately 2.23m.

The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y –axis. This can be diagrammatically represented as

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadways, 18m from the middle.

Here, AB = 30m, OC = 6m, and BC = 50m.

The equation of the parabola is of the from x2 = 4ay (as it is opening upwards).

The coordinates of point A are (50, 30 -6) = (50, 24)

Since, A(50, 24) is a point on the parabola.

(50)² = 4a(24)

⇒ a =

Thus, Equation of the parabola,

The x – coordinate of point D is 18.

Hence, at x = 18,

6(18)² = 625y

⇒ y =

⇒ y = 3.11(approx.)

Thus, DE = 3.11 m

DF = DE +EF = 3.11m +6m = 9.11m

Thus, the length of the supporting wire attached to the roadway 18m from the middle is approximately 9.11m.

Since, the height and width of the arc from the centre is 2m and 8m respectively, it is clear that the length of the major axis is 8m, while the length of the semi- minor axis is 2m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis. Hence, the semi- ellipse can be diagrammatically represented as

The equation of the semi – ellipse will be of the from ……..(1)

Let A be a point on the major axis such that AB = 1.5m.

Draw ACOB.

OA = (4 – 1.5)m = 2.5m

The x – coordinate of point C is 2.5

On substituting the value of x with 2.5 in equation (1), we get,

⇒ y² = 2.4375

⇒ y = 1.56(approx.)

Thus, AC = 1.56m

Hence, the height of the arch at a point 1.5m from one end is approximately 1.56m.

Let AB be the rod making an angle Ɵ with OX and P(x,y) be the point on it such that AP = 3cm.

Then, PB = AB – AP = (12 – 3)cm = 9cm [AB = 12cm]

From P, draw PQ OY and PROX.

In ΔPBQ,

Since, sin²Ɵ +cos²Ɵ = 1,

Or,

Thus, the equation of the locus of point P on the rod is .

The given parabola is x² = 12y.

On comparing this equation with x2 = 4ay, we get,

4a = 12

⇒ a = 3

Thus, the coordinates of foci are S(0,a) = S(0,3).

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At y = 3, x² = 12(3)

⇒ x² = 36

⇒ x = 6

Thus, the coordinates of A are (-6,3), while the coordinates of B are (6,3)

Therefore, the vertices of ΔOAB are O(0,0), A (-6,3) and B(6,3).

Area of ΔOAB = unit²

= unit²

= unit²

= unit²

unit²

=18unit²

Let A and B be the positions of the two flag posts and P(x,y) be the position of the man.

Accordingly, PA + PB = 10.

We know that if a point moves in plane in such a way that the sum of its distance from two fixed point is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described b the man is an ellipse where the length of the major axis is 10m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x- axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, 2a = 10 ⇒ a = 5

Distance between the foci (2c) = 8

⇒ c = 4

On using the relation, c = , we get,

4 =

⇒ 16 = 25 – b²

⇒ b² = 25 -1 6 = 9

⇒ b = 3

Hence, equation of the path traced by the man is

Let OAB be the equilateral triangle inscribed in parabola y² = 4ax.

Let AB intersect the x – axis at point C.

Let OC = k

From the equation of the given parabola, we have,

y² = 4ak

⇒ y = 2

Thus, the respective coordinates of points A and B are (k, 2 ), and (k, -2)

AB = CA + CB = 2

Since, OAB is an equilateral triangle, OA² = AB².

Thus,

⇒ k² + 4ak = 16ak

⇒ k² = 12ak

⇒ k = 12a

Thus, AB =

Thus, the side of the equilateral triangle inscribed in parabola y² = 4ax is .