# Find the Equation of the Hyperbola Satisfying the Given Condition: Foci (0, ± 12), Latus-rectum = 36 - Mathematics

find the equation of the hyperbola satisfying the given condition:

foci (0, ± 12), latus-rectum = 36

#### Solution

The foci of the hyperbola are $\left( 0, \pm 12 \right)$ and the latus rectum is 36.
Thus, the value of  $ae = 12$.

and $\frac{2 b^2}{a} = 36$

$\Rightarrow b^2 = 18a$

Now, using the relation

$b^2 = a^2 ( e^2 - 1)$,we get:

$\Rightarrow 18a = 144 - a^2$

$\Rightarrow a^2 + 18a - 144 = 0$

$\Rightarrow \left( a + 24 \right)\left( a - 6 \right) = 0$

$\Rightarrow a = - 24, \text { or }6$

$b^2 = - 432 \text { or }108$(but negative value is not possible)\]

Thus, the equation of the hyperbola is

$\frac{y^2}{36} - \frac{x^2}{108} = 1$.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Exercise 27.1 | Q 11.1 | Page 14