Find the equations of the hyperbola satisfying the given conditions Foci (± 4, 0), the latus rectum is of length 12

Asked by Pragya Singh | 1 year ago |  101

##### Solution :-

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

Here, the foci are on x-axis.

The equation of the hyperbola is of the form

$$\dfrac{x^2}{a^2}$$ +$$\dfrac{y^2}{b^2}$$ = 1

Since, the foci are (± 4, 0), so, c = 4

Length of latus rectum is 12

$$\dfrac{2b^2}{a}$$ = 12

2b2 = 12a

b2 = $$\dfrac{12a}{2}$$

= 6a

It is known that, a2 + b2 = c2

a2 + 6a = 16

a2 + 6a – 16 = 0

a2 + 8a – 2a – 16 = 0

(a + 8) (a – 2) = 0

a = -8 or 2

Since, a is non – negative, a = 2

So, b2 = 6a

= 6 × 2

= 12

The equation of the hyperbola is

$$\dfrac{x^2}{4}-\dfrac{y^2}{12}=1$$

Answered by Abhisek | 1 year ago

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