Learn Exercise 11.4 with Free Lessons & Tips

The given equation is.

On comparing this equation with the standard equation of hyperbola i.e.,, we obtain a = 4 and b = 3.

We know that a² + b² = c².

Therefore,

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

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

Length of latus rectum

The given equation is.

On comparing this equation with the standard equation of hyperbola i.e.,, we obtain a = 3 and.

We know that a² + b² = c².

Therefore,

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

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

Length of latus rectum

The given equation is 9y² – 4x² = 36.

It can be written as

9y² – 4x² = 36

On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain a = 2 and b = 3.

We know that a² + b² = c².

Therefore,

The coordinates of the foci are.

The coordinates of the vertices are.

Length of latus rectum

The given equation is 16x² – 9y² = 576.

It can be written as

16x² – 9y² = 576

On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain a = 6 and b = 8.

We know that a² + b² = c².

Therefore,

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

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

Length of latus rectum

The given equation is 5y² – 9x² = 36.

On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain a = and b = 2.

We know that a² + b² = c².

Therefore, the coordinates of the foci are.

The coordinates of the vertices are.

Length of latus rectum

The given equation is 49y² – 16x² = 784.

It can be written as
49y² – 16x² = 784

On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain a = 4 and b = 7.

We know that a² + b² = c².

Therefore,

The coordinates of the foci are.

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

Length of latus rectum

Vertices (±2, 0), foci (±3, 0)

Here, the vertices are on the x-axis.

Therefore, 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, the equation of the hyperbola is.

Vertices (0, ±5), foci (0, ±8)

Here, the vertices are on the y-axis.

Therefore, 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, the equation of the hyperbola is.

Vertices (0, ±3), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, 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.

∴3² + b² = 5²

⇒ b² = 25 – 9 = 16

Thus, the equation of the hyperbola is.

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

Here, the foci are on the y-axis.

Therefore, 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

Thus, the equation of the hyperbola is.

Foci, the latus rectum is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the foci are, c =.

Length of latus rectum = 8

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.

∴b² = 4a = 4 × 5 = 20

Thus, the equation of the hyperbola is.

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

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the foci are (±4, 0), c = 4.

Length of latus rectum = 12

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.

∴b² = 6a = 6 × 2 = 12

Thus, the equation of the hyperbola is.

Vertices (±7, 0),

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

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

It is given that

We know that a² + b² = c².

Thus, the equation of the hyperbola is.

Foci, passing through (2, 3)

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form.

Since the foci are, c =.

We know that a² + b² = c².

∴ a² + b² = 10

⇒ b² = 10 – a² … (1)

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

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

In hyperbola, c > a, i.e., c² > a²

∴ a² = 5

⇒ b² = 10 – a² = 10 – 5 = 5

Thus, the equation of the hyperbola is.