# Find the equation of the hyperbola satisfying the given conditions: Foci (± 3√5, 0), the latus rectum is of length 8

**Solution:**

Foci (± 3√5, 0),

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 x^{2}/a^{2} - y^{2}/b^{2} = 1

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

Since the length of latus rectum is 8,

Then,

⇒ 2b^{2}/a = 24

⇒ b^{2} = 4a

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

Hence,

⇒ a^{2} + 4a = 45

⇒ a^{2} + 4a - 45 = 0

⇒ a^{2} + 9a - 5a - 45 = 0

⇒ a (a + 9) - 5(a + 9) = 0

⇒ (a + 9)(a - 5) = 0

⇒ a = - 9, 5

Since a is non-negative, a = 5

Therefore,

⇒ b^{2} = 4a

⇒ b^{2} = 4 x 5

⇒ b^{2} = 20

Thus, the equation of the hyperbola is x^{2}/25 - y^{2}/20 = 1

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.4 Question 12

## Find the equation of the hyperbola satisfying the given conditions: Foci (± 3√5, 0), the latus rectum is of length 8

**Summary:**

The equation of the hyperbola is x^{2}/25 - y^{2}/20 = 1 while the Foci is (± 3√5, 0) and the latus rectum is of length 8

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