# Chapter – 11: Conic Sections

# EXERCISE 11.1

**In each of the following Exercise 1 to 5, find the equation of the circle with1. Centre (0, 2) and radius 2**

**Solution: –**

Given:

Centre (0, 2) and radius 2

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x – h)^{2 }+ (y – k)^{2 }= r^{2}

So, centre (h, k) = (0, 2) and radius (r) = 2

The equation of the circle is

(x – 0)^{2 }+ (y – 2)^{2 }= 2^{2}

x^{2 }+ y^{2 }+ 4 – 4y = 4

x^{2 }+ y^{2} – 4y = 0

∴ The equation of the circle is x^{2 }+ y^{2} – 4y = 0

**2. Centre (–2, 3) and radius 4**

**Solution: –**

Given:

Centre (-2, 3) and radius 4

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x – h)^{2 }+ (y – k)^{2 }= r^{2}

So, centre (h, k) = (-2, 3) and radius (r) = 4

The equation of the circle is

(x + 2)^{2} + (y – 3)^{2} = (4)^{2}

x^{2} + 4x + 4 + y^{2} – 6y + 9 = 16

x^{2} + y^{2} + 4x – 6y – 3 = 0

∴ The equation of the circle is x^{2} + y^{2} + 4x – 6y – 3 = 0

**3. Centre (1/2, 1/4) and radius (1/12)**

**Solution: –**

Given:

Centre (1/2, 1/4) and radius 1/12

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x – h)^{2 }+ (y – k)^{2 }= r^{2}

So, centre (h, k) = (1/2, 1/4) and radius (r) = 1/12

The equation of the circle is

(x – 1/2)^{2} + (y – 1/4)^{2} = (1/12)^{2}

x^{2} – x + ¼ + y^{2} – y/2 + 1/16 = 1/144

x^{2} – x + ¼ + y^{2} – y/2 + 1/16 = 1/144

144x^{2} – 144x + 36 + 144y^{2} – 72y + 9 – 1 = 0

144x^{2} – 144x + 144y^{2} – 72y + 44 = 0

36x^{2} + 36x + 36y^{2} – 18y + 11 = 0

36x^{2} + 36y^{2} – 36x – 18y + 11= 0

∴ The equation of the circle is 36x^{2} + 36y^{2} – 36x – 18y + 11= 0

**4. Centre (1, 1) and radius √2**

**Solution: –**

Given:

Centre (1, 1) and radius √2

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x – h)^{2 }+ (y – k)^{2 }= r^{2}

So, centre (h, k) = (1, 1) and radius (r) = √2

The equation of the circle is

(x-1)^{2} + (y-1)^{2 }= (√2)^{2}

x^{2} – 2x + 1 + y^{2} -2y + 1 = 2

x^{2} + y^{2} – 2x -2y = 0

∴ The equation of the circle is x^{2} + y^{2} – 2x -2y = 0

**5. Centre (–a, –b) and radius √(a**^{2}** – b**^{2}**)**

**Solution: –**

Given:

Centre (-a, -b) and radius √(a^{2} – b^{2})

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x – h)^{2 }+ (y – k)^{2 }= r^{2}

So, centre (h, k) = (-a, -b) and radius (r) = √(a^{2} – b^{2})

The equation of the circle is

(x + a)^{2} + (y + b)^{2} = (√(a^{2} – b^{2})^{2})

x^{2} + 2ax + a^{2} + y^{2} + 2by + b^{2} = a^{2} – b^{2}

x^{2} + y^{2} +2ax + 2by + 2b^{2} = 0

∴ The equation of the circle is x^{2} + y^{2} +2ax + 2by + 2b^{2} = 0

**In each of the following Exercise 6 to 9, find the centre and radius of the circles.**

6. (x + 5)^{2} + (y – 3)^{2} = 36

**Solution: –**

Given:

The equation of the given circle is (x + 5)^{2} + (y – 3)^{2} = 36

(x – (-5))^{2} + (y – 3)^{2} = 6^{2} [which is of the form (x – h)^{2} + (y – k )^{2} = r^{2}]

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

∴ The centre of the given circle is (-5, 3) and its radius is 6.

**7. x**^{2}** + y**^{2}** – 4x – 8y – 45 = 0**

**Solution: –**

Given:

The equation of the given circle is x^{2} + y^{2} – 4x – 8y – 45 = 0.

x^{2} + y^{2} – 4x – 8y – 45 = 0

(x^{2} – 4x) + (y^{2} -8y) = 45

(x^{2} – 2(x) (2) + 2^{2}) + (y^{2} – 2(y) (4) + 4^{2}) – 4 – 16 = 45

(x – 2)^{2} + (y – 4)^{2} = 65

(x – 2)^{2} + (y – 4)^{2} = (√65)^{2} [which is form (x-h)^{2} +(y-k)^{2} = r^{2}]

Where h = 2, K = 4 and r = √65

∴ The centre of the given circle is (2, 4) and its radius is √65.

**8. x**^{2}** + y**^{2}** – 8x + 10y – 12 = 0**

**Solution: –**

Given:

The equation of the given circle is x^{2} + y^{2} -8x + 10y -12 = 0.

x^{2} + y^{2} – 8x + 10y – 12 = 0

(x^{2} – 8x) + (y^{2 }+ 10y) = 12

(x^{2} – 2(x) (4) + 4^{2}) + (y^{2} – 2(y) (5) + 5^{2}) – 16 – 25 = 12

(x – 4)^{2} + (y + 5)^{2} = 53

(x – 4)^{2} + (y – (-5))^{2} = (√53)^{2} [which is form (x-h)^{2} +(y-k)^{2}= r^{2}]

Where h = 4, K= -5 and r = √53

∴ The centre of the given circle is (4, -5) and its radius is √53.

**9. 2x**^{2}** + 2y**^{2}** – x = 0**

**Solution: –**

The equation of the given of the circle is 2x^{2} + 2y^{2} –x = 0.

2x^{2} + 2y^{2} –x = 0

(2x^{2} + x) + 2y^{2} = 0

(x^{2} – 2 (x) (1/4) + (1/4)^{2}) + y^{2} – (1/4)^{2} = 0

(x – 1/4)^{2} + (y – 0)^{2} = (1/4)^{2} [which is form (x-h)^{2} +(y-k)^{2}= r^{2}]

Where, h = ¼, K = 0, and r = ¼

∴ The center of the given circle is (1/4, 0) and its radius is 1/4.

**10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.**

**Solution: –**

Let us consider the equation of the required circle be (x – h)^{2}+ (y – k)^{2} = r^{2}

We know that the circle passes through points (4,1) and (6,5)

So,

(4 – h)^{2 }+ (1 – k)^{2} = r^{2} ……………..(1)

(6– h)^{2}+ (5 – k)^{2} = r^{2} ………………(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 + h^{2} +1 -2k +k^{2} = 36 -12h +h^{2}+15 – 10k + k^{2}

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} = r^{2}

(1)^{2} + (-3)^{2} = r^{2}

1+9 = r^{2}

r = √10

so now, (x – 3)^{2 }+ (y – 4)^{2} = (√10)^{2}

x^{2} – 6x + 9 + y^{2} – 8y + 16 =10

x^{2} + y^{2} – 6x – 8y + 15 = 0

∴ The equation of the required circle is x^{2} + y^{2} – 6x – 8y + 15 = 0

**11. Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.**

**Solution: –**

Let us consider the equation of the required circle be (x – h)^{2 }+ (y – k)^{2} = r^{2}

We know that the circle passes through points (2,3) and (-1,1).

(2 – h)^{2}+ (3 – k)^{2} =r^{2} ……………..(1)

(-1 – h)^{2}+ (1– k)^{2} =r^{2} ………………(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 + h^{2} +9 -6k +k^{2} = 1 + 2h +h^{2}+1 – 2k + k^{2}

4 – 4h +9 -6k = 1 + 2h + 1 -2k

6h + 4k =11……………. (4)

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

6h+ 4k – 6(h-3k) = 11 – 66

6h + 4k – 6h + 18k = 11 – 66

22 k = – 55

K = -5/2

Substitute this value of K in equation (4) to get,

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

6h – 10 = 11

6h = 21

h = 21/6

h = 7/2

We obtain h = 7/2and k = -5/2

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

(2 – 7/2)^{2} + (3 + 5/2)^{2} = r^{2}

[(4-7)/2]^{2} + [(6+5)/2]^{2} = r^{2}

(-3/2)^{2} + (11/2)^{2} = r^{2}

9/4 + 121/4 = r^{2}

130/4 = r^{2}

The equation of the required circle is

(x – 7/2)^{2} + (y + 5/2)^{2} = 130/4

[(2x-7)/2]^{2} + [(2y+5)/2]^{2} = 130/4

4x^{2} -28x + 49 +4y^{2} + 20y + 25 =130

4x^{2} +4y^{2} -28x + 20y – 56 = 0

4(x^{2} +y^{2} -7x + 5y – 14) = 0

x^{2 }+ y^{2 }– 7x + 5y – 14 = 0

∴ The equation of the required circle is x^{2 }+ y^{2 }– 7x + 5y – 14 = 0

**12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).**

**Solution: –**

Let us consider the equation of the required circle be (x – h)^{2}+ (y – k)^{2} = r^{2}

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

So now, the equation of the circle is (x – h)^{2} + y^{2} = 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}+ 3^{2} = 25

(2 – h)^{2} = 25-9

(2 – h)^{2} = 16

2 – h = **±** **√**16 = **±** 4

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

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

Then, when h = -2, the equation of the circle becomes

(x + 2)^{2} + y^{2} = 25

x^{2} + 12x + 36 + y^{2} = 25

x^{2} + y^{2} + 4x – 21 = 0

When h = 6, the equation of the circle becomes

(x – 6)^{2} + y^{2} = 25

x^{2} -12x + 36 + y^{2} = 25

x^{2} + y^{2} -12x + 11 = 0

∴ The equation of the required circle is x^{2} + y^{2} + 4x – 21 = 0 and x^{2} + y^{2} -12x + 11 = 0

**13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.**

**Solution: –**

Let us consider the equation of the required circle be (x – h)^{2}+ (y – k)^{2} =r^{2}

We know that the circle passes through (0, 0),

So, (0 – h)^{2}+ (0 – k)^{2} = r^{2}

h^{2} + k^{2} = r^{2}

Now, The equation of the circle is (x – h)^{2 }+ (y – k)^{2} = h^{2} + k^{2}.

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

i.e., the circle passes through points (a, 0) and (0, b).

So, (a – h)^{2}+ (0 – k)^{2} =h^{2} +k^{2}……………..(1)

(0 – h)^{2}+ (b– k)^{2} =h^{2} +k^{2}………………(2)

From equation (1), we obtain

a^{2} – 2ah + h^{2} +k^{2} = h^{2} +k^{2}

a^{2} – 2ah = 0

a(a – 2h) =0

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

However, a ≠ 0; hence, (a -2h) = 0

h = a/2

From equation (2), we obtain

h^{2} – 2bk + k^{2} + b^{2}= h^{2} +k^{2}

b^{2} – 2bk = 0

b(b– 2k) = 0

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

However, a ≠ 0; hence, (b -2k) = 0

k =b/2

So, the equation is

(x – a/2)^{2} + (y – b/2)^{2} = (a/2)^{2} + (b/2)^{2}

[(2x-a)/2]^{2} + [(2y-b)/2]^{2} = (a^{2} + b^{2})/4

4x^{2} – 4ax + a^{2} +4y^{2} – 4by + b^{2} = a^{2} + b^{2}

4x^{2} + 4y^{2} -4ax – 4by = 0

4(x^{2} +y^{2} -7x + 5y – 14) = 0

x^{2 }+ y^{2 }– ax – by = 0

∴ The equation of the required circle is x^{2 }+ y^{2 }– ax – by = 0

**14. Find the equation of a circle with centre (2,2) and passes through the point (4,5).**

**Solution: –**

Given:

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

We know that 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 = **√**[(2-4)^{2} + (2-5)^{2}]

= √[(-2)^{2} + (-3)^{2}]

= √[4+9]

= √13

The equation of the circle is given as

(x– h)^{2}+ (y – k)^{2} = r^{2}

(x –h)^{2} + (y – k)^{2} = (√13)^{2}

(x –2)^{2} + (y – 2)^{2} = (√13)^{2}

x^{2} – 4x + 4 + y^{2 }– 4y + 4 = 13

x^{2} + y^{2} – 4x – 4y = 5

∴ The equation of the required circle is x^{2} + y^{2} – 4x – 4y = 5

**15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x**^{2}** + y**^{2}** = 25?**

**Solution: –**

Given:

The equation of the given circle is x^{2} +y^{2} = 25.

x^{2} + y^{2} = 25

(x – 0)^{2} + (y – 0)^{2} = 5^{2} [which is of the form (x – h)^{2 }+ (y – k)^{2} = r^{2}]

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

So the distance between point (-2.5, 3.5) and the centre (0,0) is

= √[(-2.5 – 0)^{2} + (-3.5 – 0)^{2}]

= √(6.25 + 12.25)

= √18.5

= 4.3 [which is < 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.

# EXERCISE 11.2

**In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.**

1. y^{2}** = 12x**

**Solution: –**

Given:

The equation is y^{2} = 12x

Here we know that the coefficient of x is positive.

So, the parabola opens towards the right.

On comparing this equation with y^{2} = 4ax, we get,

4a = 12

a = 3

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

Since, the given equation involves y^{2}, the axis of the parabola is the x-axis.

∴ The equation of directrix, x = -a, then,

x + 3 = 0

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

**2. x**^{2}** = 6y**

**Solution: –**

Given:

The equation is x^{2} = 6y

Here we know that the coefficient of y is positive.

So, the parabola opens upwards.

On comparing this equation with x^{2} = 4ay, we get,

4a = 6

a = 6/4

= 3/2

Thus, the co-ordinates of the focus = (0,a) = (0, 3/2)

Since, the given equation involves x^{2}, the axis of the parabola is the y-axis.

∴ The equation of directrix, y =-a, then,

y = -3/2

Length of latus rectum = 4a = 4(3/2) = 6

**3. y**^{2}** = – 8x**

**Solution: –**

Given:

The equation is y^{2} = -8x

Here we know that the coefficient of x is negative.

So, the parabola open towards the left.

On comparing this equation with y^{2} = -4ax, we get,

-4a = -8

a = -8/-4 = 2

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

Since, the given equation involves y^{2}, the axis of the parabola is the x-axis.

∴ Equation of directrix, x =a, then,

x = 2

Length of latus rectum = 4a = 4 (2) = 8

**4. x**^{2}** = – 16y**

**Solution: –**

Given:

The equation is x^{2} = -16y

Here we know that the coefficient of y is negative.

So, the parabola opens downwards.

On comparing this equation with x^{2} = -4ay, we get,

-4a = -16

a = -16/-4

= 4

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

Since, the given equation involves x^{2}, the axis of the parabola is the y-axis.

∴ The equation of directrix, y =a, then,

y = 4

Length of latus rectum = 4a = 4(4) = 16

**5. y**^{2}** = 10x**

**Solution: –**

Given:

The equation is y^{2} = 10x

Here we know that the coefficient of x is positive.

So, the parabola open towards the right.

On comparing this equation with y^{2} = 4ax, we get,

4a = 10

a = 10/4 = 5/2

Thus, co-ordinates of the focus = (a,0) = (5/2, 0)

Since, the given equation involves y^{2}, the axis of the parabola is the x-axis.

∴ The equation of directrix, x =-a, then,

x = – 5/2

Length of latus rectum = 4a = 4(5/2) = 10

**6. x**^{2}** = – 9y**

**Solution: –**

Given:

The equation is x^{2} = -9y

Here we know that the coefficient of y is negative.

So, the parabola open downwards.

On comparing this equation with x^{2} = -4ay, we get,

-4a = -9

a = -9/-4 = 9/4

Thus, co-ordinates of the focus = (0,-a) = (0, -9/4)

Since, the given equation involves x^{2}, the axis of the parabola is the y-axis.

∴ The equation of directrix, y = a, then,

y = 9/4

Length of latus rectum = 4a = 4(9/4) = 9

**In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:**

7. Focus (6,0); directrix x = – 6

**Solution: –**

Given:

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

We know that the focus lies on the x–axis is the axis of the parabola.

So, the equation of the parabola is either of the form y^{2} = 4ax or y^{2} = -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 y^{2} = 4ax.

Here, a = 6

∴ The equation of the parabola is y^{2} = 24x.

**8. Focus (0,–3); directrix y = 3**

**Solution: –**

Given:

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

We know that the focus lies on the y–axis, the y-axis is the axis of the parabola.

So, the equation of the parabola is either of the form x^{2} = 4ay or x^{2} = -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 x^{2} = -4ay.

Here, a = 3

∴ The equation of the parabola is x^{2} = -12y.

**9. Vertex (0, 0); focus (3, 0)**

**Solution: –**

Given:

Vertex (0, 0) and focus (3, 0)

We know that 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.]

The equation of the parabola is of the form y^{2 }= 4ax.

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

∴ The equation of the parabola is y^{2} = 4 × 3 × x,

y^{2} = 12x

**10. Vertex (0, 0); focus (–2, 0)**

**Solution: –**

Given:

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

We know that 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.]

The equation of the parabola is of the form y^{2}=-4ax.

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

∴ The equation of the parabola is y^{2} = -4 × 2 × x,

y^{2} = -8x

**11. Vertex (0, 0) passing through (2, 3) and axis is along x-axis.**

**Solution: –**

We know that 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 y^{2 }= 4ax or y^{2} = -4ax.

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

So, the equation of the parabola is of the form y^{2} = 4ax, while point (2, 3) must satisfy the equation y^{2} = 4ax.

Then,

3^{2} = 4a(2)

3^{2} = 8a

9 = 8a

a = 9/8

Thus, the equation of the parabola is

y^{2} = 4 (9/8)x

= 9x/2

2y^{2} = 9x

∴ The equation of the parabola is 2y^{2} = 9x

**12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.**

**Solution: –**

We know that the vertex is (0, 0) and the parabola is symmetric about the y-axis.

The equation of the parabola is either of the from x^{2 }= 4ay or x^{2} = -4ay.

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

So, the equation of the parabola is of the form x^{2} = 4ay, while point (5, 2) must satisfy the equation x^{2} = 4ay.

Then,

5^{2} = 4a(2)

25 = 8a

a = 25/8

Thus, the equation of the parabola is

x^{2} = 4 (25/8)y

x^{2} = 25y/2

2x^{2} = 25y

∴ The equation of the parabola is 2x^{2} = 25y

## EXERCISE 11.3

**In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.**

**1. x**^{2}**/36 + y**^{2}**/16 = 1**

**Solution: –**

Given:

The equation is x^{2}/36 + y^{2}/16 = 1

Here, the denominator of x^{2}/36 is greater than the denominator of y^{2}/16.

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

On comparing the given equation with x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a = 6 and b = 4.

c = √(a^{2} – b^{2})

= √(36-16)

= √20

= 2√5

Then,

The coordinates of the foci are (2√5, 0) and (-2√5, 0).

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

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (4) = 8

Eccentricity, e = c/a = 2√5/6 = √5/3

Length of latus rectum = 2b^{2}/a = (2×16)/6 = 16/3

**2. x**^{2}**/4 + y**^{2}**/25 = 1**

**Solution: –**

Given:

The equation is x^{2}/4 + y^{2}/25 = 1

Here, the denominator of y^{2}/25 is greater than the denominator of x^{2}/4.

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

On comparing the given equation with x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a = 5 and b = 2.

c = √(a^{2} – b^{2})

= √(25-4)

= √21

Then,

The coordinates of the foci are (0, √21) and (0, -√21).

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

Length of major axis = 2a = 2 (5) = 10

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = √21/5

Length of latus rectum = 2b^{2}/a = (2×2^{2})/5 = (2×4)/5 = 8/5

**3. x**^{2}**/16 + y**^{2}**/9 = 1**

**Solution: –**

Given:

The equation is x^{2}/16 + y^{2}/9 = 1 or x^{2}/4^{2} + y^{2}/3^{2} = 1

Here, the denominator of x^{2}/16 is greater than the denominator of y^{2}/9.

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

On comparing the given equation with x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a = 4 and b = 3.

c = √(a^{2} – b^{2})

= √(16-9)

= √7

Then,

The coordinates of the foci are (√7, 0) and (-√7, 0).

The coordinates of the vertices are (4, 0) and (-4, 0)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (3) = 6

Eccentricity, e = c/a = √7/4

Length of latus rectum = 2b^{2}/a = (2×3^{2})/4 = (2×9)/4 = 18/4 = 9/2

**4. x**^{2}**/25 + y**^{2}**/100 = 1**

**Solution: –**

Given:

The equation is x^{2}/25 + y^{2}/100 = 1

Here, the denominator of y^{2}/100 is greater than the denominator of x^{2}/25.

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

On comparing the given equation with x^{2}/b^{2} + y^{2}/a^{2} = 1, we get

b = 5 and a =10.

c = √(a^{2} – b^{2})

= √(100-25)

= √75

= 5√3

Then,

The coordinates of the foci are (0, 5√3) and (0, -5√3).

The coordinates of the vertices are (0, √10) and (0, -√10)

Length of major axis = 2a = 2 (10) = 20

Length of minor axis = 2b = 2 (5) = 10

Eccentricity, e = c/a = 5√3/10 = √3/2

Length of latus rectum = 2b^{2}/a = (2×5^{2})/10 = (2×25)/10 = 5

**5. x**^{2}**/49 + y**^{2}**/36 = 1**

**Solution: –**

Given:

The equation is x^{2}/49 + y^{2}/36 = 1

Here, the denominator of x^{2}/49 is greater than the denominator of y^{2}/36.

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

On comparing the given equation with x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

b = 6 and a =7

c = √(a^{2} – b^{2})

= √(49-36)

= √13

Then,

The coordinates of the foci are (√13, 0) and (-√3, 0).

The coordinates of the vertices are (7, 0) and (-7, 0)

Length of major axis = 2a = 2 (7) = 14

Length of minor axis = 2b = 2 (6) = 12

Eccentricity, e = c/a = √13/7

Length of latus rectum = 2b^{2}/a = (2×6^{2})/7 = (2×36)/7 = 72/7

**6. x**^{2}**/100 + y**^{2}**/400 = 1**

**Solution: –**

Given:

The equation is x^{2}/100 + y^{2}/400 = 1

Here, the denominator of y^{2}/400 is greater than the denominator of x^{2}/100.

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

On comparing the given equation with x^{2}/b^{2} + y^{2}/a^{2} = 1, we get

b = 10 and a =20.

c = √(a^{2} – b^{2})

= √(400-100)

= √300

= 10√3

Then,

The coordinates of the foci are (0, 10√3) and (0, -10√3).

The coordinates of the vertices are (0, 20) and (0, -20)

Length of major axis = 2a = 2 (20) = 40

Length of minor axis = 2b = 2 (10) = 20

Eccentricity, e = c/a = 10√3/20 = √3/2

Length of latus rectum = 2b^{2}/a = (2×10^{2})/20 = (2×100)/20 = 10

**7. 36x**^{2}** + 4y**^{2}** = 144**

**Solution: –**

Given:

The equation is 36x^{2} + 4y^{2} = 144 or x^{2}/4 + y^{2}/36 = 1 or x^{2}/2^{2} + y^{2}/6^{2} = 1

Here, the denominator of y^{2}/6^{2} is greater than the denominator of x^{2}/2^{2}.

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

On comparing the given equation with x^{2}/b^{2} + y^{2}/a^{2} = 1, we get

b = 2 and a = 6.

c = √(a^{2} – b^{2})

= √(36-4)

= √32

= 4√2

Then,

The coordinates of the foci are (0, 4√2) and (0, -4√2).

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

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = 4√2/6 = 2√2/3

Length of latus rectum = 2b^{2}/a = (2×2^{2})/6 = (2×4)/6 = 4/3

**8. 16x**^{2}** + y**^{2}** = 16**

**Solution: –**

Given:

The equation is 16x^{2} + y^{2} = 16 or x^{2}/1 + y^{2}/16 = 1 or x^{2}/1^{2}+ y^{2}/4^{2} = 1

Here, the denominator of y^{2}/4^{2} is greater than the denominator of x^{2}/1^{2}.

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

On comparing the given equation with x^{2}/b^{2} + y^{2}/a^{2} = 1, we get

b =1 and a =4.

c = √(a^{2} – b^{2})

= √(16-1)

= √15

Then,

The coordinates of the foci are (0, √15) and (0, -√15).

The coordinates of the vertices are (0, 4) and (0, -4)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (1) = 2

Eccentricity, e = c/a = √15/4

Length of latus rectum = 2b^{2}/a = (2×1^{2})/4 = 2/4 = ½

**9. 4x**^{2}** + 9y**^{2}** = 36**

**Solution: –**

Given:

The equation is 4x^{2} + 9y^{2} = 36 or x^{2}/9 + y^{2}/4 = 1 or x^{2}/3^{2}+ y^{2}/2^{2} = 1

Here, the denominator of x^{2}/3^{2} is greater than the denominator of y^{2}/2^{2}.

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

On comparing the given equation with x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a =3 and b =2.

c = √(a^{2} – b^{2})

= √(9-4)

= √5

Then,

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

The coordinates of the vertices are (3, 0) and (-3, 0)

Length of major axis = 2a = 2 (3) = 6

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = √5/3

Length of latus rectum = 2b^{2}/a = (2×2^{2})/3 = (2×4)/3 = 8/3

**In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:**

**10. Vertices (± 5, 0), foci (± 4, 0)**

**Solution: –**

Given:

Vertices (± 5, 0) and foci (± 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the form x^{2}/a^{2} + y^{2}/b^{2} = 1, where ‘a’ is the semi-major axis.

Then, a = 5 and c = 4.

It is known that a^{2} = b^{2 }+ c^{2}.

So, 5^{2} = b^{2 }+ 4^{2}

25 = b^{2} + 16

b^{2} = 25 – 16

b = √9

= 3

∴ The equation of the ellipse is x^{2}/5^{2} + y^{2}/3^{2} = 1 or x^{2}/25 + y^{2}/9 = 1

**11. Vertices (0, ± 13), foci (0, ± 5)**

**Solution: –**

Given:

Vertices (0, ± 13) and foci (0, ± 5)

Here, the vertices are on the y-axis.

So, the equation of the ellipse will be of the form x^{2}/b^{2} + y^{2}/a^{2} = 1, where ‘a’ is the semi-major axis.

Then, a =13 and c = 5.

It is known that a^{2} = b^{2 }+ c^{2}.

13^{2} = b^{2}+5^{2}

169 = b^{2} + 15

b^{2} = 169 – 125

b = √144

= 12

∴ The equation of the ellipse is x^{2}/12^{2} + y^{2}/13^{2} = 1 or x^{2}/144 + y^{2}/169 = 1

**12. Vertices (± 6, 0), foci (± 4, 0)**

**Solution: –**

Given:

Vertices (± 6, 0) and foci (± 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the form x^{2}/a^{2} + y^{2}/b^{2} = 1, where ‘a’ is the semi-major axis.

Then, a = 6 and c = 4.

It is known that a^{2} = b^{2 }+ c^{2}.

6^{2} = b^{2}+4^{2}

36 = b^{2} + 16

b^{2} = 36 – 16

b = √20

∴ The equation of the ellipse is x^{2}/6^{2} + y^{2}/(√20)^{2} = 1 or x^{2}/36 + y^{2}/20 = 1

**13. Ends of major axis (± 3, 0), ends of minor axis (0, ±2)**

**Solution: –**

Given:

Ends of major axis (± 3, 0) and ends of minor axis (0, ±2)

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

So, the equation of the ellipse will be of the form x^{2}/a^{2} + y^{2}/b^{2} = 1, where ‘a’ is the semi-major axis.

Then, a = 3 and b = 2.

∴ The equation for the ellipse x^{2}/3^{2} + y^{2}/2^{2} = 1 or x^{2}/9 + y^{2}/4 = 1

**14. Ends of major axis (0, ±√5), ends of minor axis (±1, 0)**

**Solution: –**

Given:

Ends of major axis (0, ±√5) and ends of minor axis (±1, 0)

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

So, the equation of the ellipse will be of the form x^{2}/b^{2} + y^{2}/a^{2} = 1, where ‘a’ is the semi-major axis.

Then, a = √5 and b = 1.

∴ The equation for the ellipse x^{2}/1^{2} + y^{2}/(√5)^{2} = 1 or x^{2}/1 + y^{2}/5 = 1

**15. Length of major axis 26, foci (±5, 0)**

**Solution: –**

Given:

Length of major axis is 26 and foci (±5, 0)

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

^{2}/a^{2} + y^{2}/b^{2} = 1, where ‘a’ is the semi-major axis.

Then, 2a = 26

a = 13 and c = 5.

It is known that a^{2} = b^{2 }+ c^{2}.

13^{2} = b^{2}+5^{2}

169 = b^{2} + 25

b^{2} = 169 – 25

b = √144

= 12

∴ The equation of the ellipse is x^{2}/13^{2} + y^{2}/12^{2} = 1 or x^{2}/169 + y^{2}/144 = 1

**16. Length of minor axis 16, foci (0, ±6).**

**Solution: –**

Given:

Length of minor axis is 16 and foci (0, ±6).

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

So, the equation of the ellipse will be of the form x^{2}/b^{2} + y^{2}/a^{2} = 1, where ‘a’ is the semi-major axis.

Then, 2b =16

b = 8 and c = 6.

It is known that a^{2} = b^{2 }+ c^{2}.

a^{2} = 8^{2 }+ 6^{2}

= 64 + 36

=100

a = √100

= 10

∴ The equation of the ellipse is x^{2}/8^{2} + y^{2}/10^{2} =1 or x^{2}/64 + y^{2}/100 = 1

**17. Foci (±3, 0), a = 4**

**Solution: –**

Given:

Foci (±3, 0) and a = 4

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

^{2}/a^{2} + y^{2}/b^{2} = 1, where ‘a’ is the semi-major axis.

Then, c = 3 and a = 4.

It is known that a^{2} = b^{2 }+ c^{2}.

a^{2} = 8^{2 }+ 6^{2}

= 64 + 36

= 100

16 = b^{2} + 9

b^{2} = 16 – 9

= 7

∴ The equation of the ellipse is x^{2}/16 + y^{2}/7 = 1

**18. b = 3, c = 4, centre at the origin; foci on the x axis.**

**Solution: –**

Given:

b = 3, c = 4, centre at the origin and foci on the x axis.

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

^{2}/a^{2} + y^{2}/b^{2} = 1, where ‘a’ is the semi-major axis.

Then, b = 3 and c = 4.

It is known that a^{2} = b^{2 }+ c^{2}.

a^{2} = 3^{2 }+ 4^{2}

= 9 + 16

=25

a = √25

= 5

∴ The equation of the ellipse is x^{2}/5^{2} + y^{2}/3^{2} or x^{2}/25 + y^{2}/9 = 1

**19. Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).**

**Solution: –**

Given:

Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Since the centre is at (0, 0) and the major axis is on the y- axis, the equation of the ellipse will be of the form x^{2}/b^{2} + y^{2}/a^{2} = 1, where ‘a’ is the semi-major axis.

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

So, by putting the values x = 3 and y = 2, we get,

3^{2}/b^{2} + 2^{2}/a^{2} = 1

9/b^{2} + 4/a^{2}…. (1)

And by putting the values x = 1 and y = 6, we get,

1^{1}/b^{2} + 6^{2}/a^{2} = 1

1/b^{2} + 36/a^{2} = 1 …. (2)

On solving equation (1) and (2), we get

b^{2} = 10 and a^{2} = 40.

∴ The equation of the ellipse is x^{2}/10 + y^{2}/40 = 1 or 4x^{2} + y ^{2} = 40

**20. Major axis on the x-axis and passes through the points (4,3) and (6,2).**

**Solution: –**

Given:

Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

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

x^{2}/a^{2} + y^{2}/b^{2} = 1…. (1) [Where ‘a’ is the semi-major axis.]

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

So by putting the values x = 4 and y = 3 in equation (1), we get,

16/a^{2} + 9/b^{2} = 1 …. (2)

Putting, x = 6 and y = 2 in equation (1), we get,

36/a^{2} + 4/b^{2} = 1 …. (3)

From equation (2)

16/a^{2} = 1 – 9/b^{2}

1/a^{2} = (1/16 (1 – 9/b^{2})) …. (4)

Substituting the value of 1/a^{2} in equation (3) we get,

36/a^{2} + 4/b^{2} = 1

36(1/a^{2}) + 4/b^{2} = 1

36[1/16 (1 – 9/b^{2})] + 4/b^{2} = 1

36/16 (1 – 9/b^{2}) + 4/b^{2} = 1

9/4 (1 – 9/b^{2}) + 4/b^{2} = 1

9/4 – 81/4b^{2} + 4/b^{2} = 1

-81/4b^{2} + 4/b^{2} = 1 – 9/4

(-81+16)/4b^{2} = (4-9)/4

-65/4b^{2} = -5/4

-5/4(13/b^{2}) = -5/4

13/b^{2} = 1

1/b^{2} = 1/13

b^{2} = 13

Now substitute the value of b^{2} in equation (4) we get,

1/a^{2} = 1/16(1 – 9/b^{2})

= 1/16(1 – 9/13)

= 1/16((13-9)/13)

= 1/16(4/13)

= 1/52

a^{2} = 52

Equation of ellipse is x^{2}/a^{2} + y^{2}/b^{2} = 1

By substituting the values of a^{2} and b^{2} in above equation we get,

x^{2}/52 + y^{2}/13 = 1

# EXERCISE 11.4

**In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.**

**1. x**^{2}**/16 – y**^{2}**/9 = 1**

**Solution: –**

Given:

The equation is x^{2}/16 – y^{2}/9 = 1 or x^{2}/4^{2} – y^{2}/3^{2} = 1

On comparing this equation with the standard equation of hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1,

We get a = 4 and b = 3,

It is known that, a^{2} + b^{2} = c^{2}

So,

c^{2} = 4^{2} + 3^{2}

=** √**25

c = 5

Then,

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

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

Eccentricity, e = c/a = 5/4

Length of latus rectum = 2b^{2}/a = (2 × 3^{2})/4 = (2×9)/4 = 18/4 = 9/2

**2. y**^{2}**/9 – x**^{2}**/27 = 1**

**Solution: –**

Given:

The equation is y^{2}/9 – x^{2}/27 = 1 or y^{2}/3^{2} – x^{2}/27^{2} = 1

On comparing this equation with the standard equation of hyperbola y^{2}/a^{2} – x^{2}/b^{2} = 1,

We get a = 3 and b = **√**27,

It is known that, a^{2} + b^{2} = c^{2}

So,

c^{2} = 3^{2} + (**√**27)^{2}

= 9 + 27

c^{2} = 36

c = **√**36

= 6

Then,

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

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

Eccentricity, e = c/a = 6/3 = 2

Length of latus rectum = 2b^{2}/a = (2 × 27)/3 = (54)/3 = 18

**3. 9y**^{2}** – 4x**^{2}** = 36**

**Solution: –**

Given:

The equation is 9y^{2} – 4x^{2} = 36 or y^{2}/4 – x^{2}/9 = 1 or y^{2}/2^{2}– x^{2}/3^{2} = 1

On comparing this equation with the standard equation of hyperbola y^{2}/a^{2} – x^{2}/b^{2} = 1,

We get a = 2 and b = 3,

It is known that, a^{2} + b^{2} = c^{2}

So,

c^{2} = 4 + 9

c^{2} = 13

c = **√**13

Then,

The coordinates of the foci are (0, **√**13) and (0, –**√**13).

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

Eccentricity, e = c/a = **√**13/2

Length of latus rectum = 2b^{2}/a = (2 × 3^{2})/2 = (2×9)/2 = 18/2 = 9

**4. 16x**^{2}** – 9y**^{2}** = 576**

**Solution: –**

Given:

The equation is 16x^{2} – 9y^{2} = 576

Let us divide the whole equation by 576, we get

16x^{2}/576 – 9y^{2}/576 = 576/576

x^{2}/36 – y^{2}/64 = 1

On comparing this equation with the standard equation of hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1,

We get a = 6 and b = 8,

It is known that, a^{2} + b^{2} = c^{2}

So,

c^{2} = 36 + 64

c^{2} = **√**100

c = 10

Then,

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

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

Eccentricity, e = c/a = 10/6 = 5/3

Length of latus rectum = 2b^{2}/a = (2 × 8^{2})/6 = (2×64)/6 = 64/3

**5. 5y**^{2}** – 9x**^{2}** = 36**

**Solution: –**

Given:

The equation is 5y^{2} – 9x^{2} = 36

Let us divide the whole equation by 36, we get

5y^{2}/36 – 9x^{2}/36 = 36/36

y^{2}/(36/5) – x^{2}/4 = 1

On comparing this equation with the standard equation of hyperbola y^{2}/a^{2} – x^{2}/b^{2} = 1,

We get a = 6/**√**5 and b = 2,

It is known that, a^{2} + b^{2} = c^{2}

So,

c^{2} = 36/5 + 4

c^{2} = 56/5

c = **√**(56/5)

= 2**√**14/**√**5

Then,

The coordinates of the foci are (0, 2**√**14/**√**5) and (0, – 2**√**14/**√**5).

The coordinates of the vertices are (0, 6/**√**5) and (0, -6/**√**5).

Eccentricity, e = c/a = (2**√**14/**√**5) / (6/**√**5) = **√**14/3

Length of latus rectum = 2b^{2}/a = (2 × 2^{2})/6/√5 = (2×4)/6/**√**5 = 4**√**5/3

**6. 49y**^{2}** – 16x**^{2}** = 784.**

**Solution: –**

Given:

The equation is 49y^{2} – 16x^{2} = 784.

Let us divide the whole equation by 784, we get

49y^{2}/784 – 16x^{2}/784 = 784/784

y^{2}/16 – x^{2}/49 = 1

^{2}/a^{2} – x^{2}/b^{2} = 1,

We get a = 4 and b = 7,

It is know that, a^{2} + b^{2} = c^{2}

So,

c^{2} = 16 + 49

c^{2} = 65

c = **√**65

Then,

The coordinates of the foci are (0, **√**65) and (0, –**√**65).

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

Eccentricity, e = c/a = **√**65/4

Length of latus rectum = 2b^{2}/a = (2 × 7^{2})/4 = (2×49)/4 = 49/2

**In each Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions**

**7. Vertices (±2, 0), foci (±3, 0)**

**Solution: –**

Given:

Vertices (±2, 0) and foci (±3, 0)

Here, the vertices are on the x-axis.

So, the equation of the hyperbola is of the form x^{2}/a^{2} – y^{2}/b^{2} = 1

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

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

It is know that, a^{2} + b^{2} = c^{2}

So, 2^{2} + b^{2} = 3^{2}

b^{2} = 9 – 4 = 5

∴ The equation of the hyperbola is x^{2}/4 – y^{2}/5 = 1

**8. Vertices (0, ± 5), foci (0, ± 8)**

**Solution: –**

Given:

Vertices (0, ± 5) and foci (0, ± 8)

Here, the vertices are on the y-axis.

So, the equation of the hyperbola is of the form y^{2}/a^{2} – x^{2}/b^{2} = 1

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

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

It is know that, a^{2} + b^{2} = c^{2}

So, 5^{2} + b^{2} = 8^{2}

b^{2} = 64 – 25 = 39

∴ The equation of the hyperbola is y^{2}/25 – x^{2}/39 = 1

**9. Vertices (0, ± 3), foci (0, ± 5)**

**Solution: –**

Given:

Vertices (0, ± 3) and foci (0, ± 5)

Here, the vertices are on the y-axis.

So, the equation of the hyperbola is of the form y^{2}/a^{2} – x^{2}/b^{2} = 1

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

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

It is known that, a^{2} + b^{2} = c^{2}

So, 3^{2} + b^{2} = 5^{2}

b^{2} = 25 – 9 = 16

∴ The equation of the hyperbola is y^{2}/9 – x^{2}/16 = 1

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

**Solution: –**

Given:

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

Here, the foci are on x-axis.

The equation of the hyperbola is of the form x^{2}/a^{2} – y^{2}/b^{2} = 1

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

Since, the length of the transverse axis is 8,

2a = 8

a = 8/2

= 4

It is known that, a^{2} + b^{2} = c^{2}

4^{2} + b^{2} = 5^{2}

b^{2} = 25 – 16

= 9

∴ The equation of the hyperbola is x^{2}/16 – y^{2}/9 = 1

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

**Solution: –**

Given:

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

Here, the foci are on y-axis.

The equation of the hyperbola is of the form y^{2}/a^{2} – x^{2}/b^{2} = 1

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

Since, the length of the conjugate axis is 24,

2b = 24

b = 24/2

= 12

It is known that, a^{2} + b^{2} = c^{2}

a^{2} + 12^{2} = 13^{2}

a^{2} = 169 – 144

= 25

∴ The equation of the hyperbola is y^{2}/25 – x^{2}/144 = 1

**12. Foci (± 3√5, 0), the latus rectum is of length 8.**

**Solution: –**

Given:

Foci (± 3√5, 0) and the latus rectum is of length 8.

Here, the foci are on x-axis.

The equation of the hyperbola is of the form x^{2}/a^{2} – y^{2}/b^{2} = 1

Since, the foci are (± 3√5, 0), so, c = ± 3√5

Length of latus rectum is 8

2b^{2}/a = 8

2b^{2} = 8a

b^{2} = 8a/2

= 4a

It is known that, a^{2} + b^{2} = c^{2}

a^{2} + 4a = 45

a^{2} + 4a – 45 = 0

a^{2} + 9a – 5a – 45 = 0

(a + 9) (a -5) = 0

a = -9 or 5

Since, a is non – negative, a = 5

So, b^{2} = 4a

= 4 × 5

= 20

∴ The equation of the hyperbola is x^{2}/25 – y^{2}/20 = 1

**13. Foci (± 4, 0), the latus rectum is of length 12**

**Solution: –**

Given:

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

Here, the foci are on x-axis.

The equation of the hyperbola is of the form x^{2}/a^{2} – y^{2}/b^{2} = 1

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

Length of latus rectum is 12

2b^{2}/a = 12

2b^{2} = 12a

b^{2} = 12a/2

= 6a

It is known that, a^{2} + b^{2} = c^{2}

a^{2} + 6a = 16

a^{2} + 6a – 16 = 0

a^{2} + 8a – 2a – 16 = 0

(a + 8) (a – 2) = 0

a = -8 or 2

Since, a is non – negative, a = 2

So, b^{2} = 6a

= 6 × 2

= 12

∴ The equation of the hyperbola is x^{2}/4 – y^{2}/12 = 1

**14. Vertices (±7, 0), e = 4/3**

**Solution: –**

Given:

Vertices (±7, 0) and e = 4/3

Here, the vertices are on the x- axis

The equation of the hyperbola is of the form x^{2}/a^{2} – y^{2}/b^{2} = 1

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

It is given that e = 4/3

c/a = 4/3

3c = 4a

Substitute the value of a, we get

3c = 4(7)

c = 28/3

It is known that, a^{2} + b^{2} = c^{2}

7^{2} + b^{2} = (28/3)^{2}

b^{2} = 784/9 – 49

= (784 – 441)/9

= 343/9

∴ The equation of the hyperbola is x^{2}/49 – 9y^{2}/343 = 1

**15. Foci (0, ±√10), passing through (2, 3)**

**Solution: –**

Given:

Foci (0, ±√10) and passing through (2, 3)

Here, the foci are on y-axis.

The equation of the hyperbola is of the form y^{2}/a^{2} – x^{2}/b^{2} = 1

Since, the foci are (±√10, 0), so, c = √10

It is known that, a^{2} + b^{2} = c^{2}

b^{2} = 10 – a^{2} ………….. (1)

It is given that the hyperbola passes through point (2, 3)

So, 9/a^{2} – 4/b^{2} = 1 … (2)

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

9/a^{2} – 4/(10-a^{2}) = 1

9(10 – a^{2}) – 4a^{2} = a^{2}(10 –a^{2})

90 – 9a^{2} – 4a^{2} = 10a^{2} – a^{4}

a^{4} – 23a^{2} + 90 = 0

a^{4} – 18a^{2} – 5a^{2 }+ 90 = 0

a^{2}(a^{2} -18) -5(a^{2} -18) = 0

(a^{2} – 18) (a^{2} -5) = 0

a^{2} = 18 or 5

In hyperbola, c > a i.e., c^{2 }> a^{2}

So, a^{2} = 5

b^{2} = 10 – a^{2}

= 10 – 5

= 5

∴ The equation of the hyperbola is y^{2}/5 – x^{2}/5 = 1.

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