Chapter 11: Conic Sections Exercise 11.4

Given:

The equation is x²/16 – y²/9 = 1 or x²/4² – y²/3² = 1

On comparing this equation with the standard equation of hyperbola x²/a² – y²/b² = 1,

We get a = 4 and b = 3,

It is known that a² + b² = c²

So,

c² = 4² + 3²

= √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 = 2b²/a = (2 × 3²)/4 = (2×9)/4 = 18/4 = 9/2

Given:

The equation is y²/9 – x²/27 = 1 or y²/3² – x²/27² = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1,

We get a = 3 and b = √27,

It is known that a² + b² = c²

So,

c² = 3² + (√27)²

= 9 + 27

c² = 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 = 2b²/a = (2 × 27)/3 = (54)/3 = 18

Given:

The equation is 9y² – 4x² = 36 or y²/4 – x²/9 = 1 or y²/2² – x²/3² = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1,

We get a = 2 and b = 3,

It is known that a² + b² = c²

So,

c² = 4 + 9

c² = 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 = 2b²/a = (2 × 3²)/2 = (2×9)/2 = 18/2 = 9

Given:

The equation is 16x² – 9y² = 576

Let us divide the whole equation by 576, we get

16x²/576 – 9y²/576 = 576/576

x²/36 – y²/64 = 1

On comparing this equation with the standard equation of hyperbola x²/a² – y²/b² = 1,

We get a = 6 and b = 8,

It is known that a² + b² = c²

So,

c² = 36 + 64

c² = √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 = 2b²/a = (2 × 8²)/6 = (2×64)/6 = 64/3

Given:

The equation is 5y² – 9x² = 36

Let us divide the whole equation by 36. We get

5y²/36 – 9x²/36 = 36/36

y²/(36/5) – x²/4 = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1,

We get a = 6/√5 and b = 2,

It is known that a² + b² = c²

So,

c² = 36/5 + 4

c² = 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 = 2b²/a = (2 × 2²)/6/√5 = (2×4)/6/√5 = 4√5/3

Given:

The equation is 49y² – 16x² = 784.

Let us divide the whole equation by 784. We get

49y²/784 – 16x²/784 = 784/784

y²/16 – x²/49 = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1,

We get a = 4 and b = 7,

It is known that a² + b² = c²

So,

c² = 16 + 49

c² = 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 = 2b²/a = (2 × 7²)/4 = (2×49)/4 = 49/2

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

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 x²/a² – y²/b² = 1

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

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

It is known that a² + b² = c²

So, 2² + b² = 3²

b² = 9 – 4 = 5

∴ The equation of the hyperbola is x²/4 – y²/5 = 1

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 y²/a² – x²/b² = 1

Since the vertices are (0, ±5), a = 5

Since the foci are (0, ±8), c = 8

It is known that a² + b² = c²

So, 5² + b² = 8²

b² = 64 – 25 = 39

∴ The equation of the hyperbola is y²/25 – x²/39 = 1

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 y²/a² – x²/b² = 1

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

Since the foci are (0, ±5), so, c = 5

It is known that a² + b² = c²

So, 3² + b² = 5²

b² = 25 – 9 = 16

∴ The equation of the hyperbola is y²/9 – x²/16 = 1

Given:

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

Here, the foci are on the x-axis.

The equation of the hyperbola is of the form x²/a² – y²/b² = 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 a² + b² = c²

4² + b² = 5²

b² = 25 – 16

= 9

∴ The equation of the hyperbola is x²/16 – y²/9 = 1