Question

find the equation of hyperbola whose foci are (0; +-12) and the length of the latus rectum is 36​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given ,

Foci of hyperbola = ( 0 , ± 12)

Length of the latus rectum = 36

So , c = 12

We know that , the length of latus rectum is given by

$\boxed{ \tt{Length \: of \: latus \: rectum = \frac{2 {(b)}^{2} }{a} }}$

Thus ,

36 = 2(b)²/a

36a = 2(b)²

(b)² = 18a --- (i)

Now , the relationship between c , a and b in hyperbola is given by

$\boxed{ \tt{{(c)}^{2} = {(a)}^{2} + {(b)}^{2} }}$

Thus ,

(12)² = (a)² + 18a

144 = (a)² + 18a

(a)² + 18a - 144 = 0

(a)² + 24a - 6a - 144 = 0

a(a + 24) - 6(a + 24) = 0

(a - 6)(a + 24) = 0

a = 6 or a = -24 { a can't be -ve }

Put the value of a in eq (i) , we get

(b)² = 18 × 6 = 108

Since , foci on the y axis

Therefore , the equation of hyperbola will be

$\boxed{ \tt{ \frac{ {(y)}^{2} }{36} - \frac{ {(x)}^{2} }{108} = 1} }$