Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find the Equation of the Hyperbola Satisfying the Given Condition: Foci (± 4, 0), the Latus-rectum = 12 - Mathematics

(vii)  find the equation of the hyperbola satisfying the given condition:

foci (± 4, 0), the latus-rectum = 12

#### Solution

The foci of the hyperbola are $\left( \pm 4, 0 \right)$ and the latus rectum is 12.
Thus, the value of  $ae = 4$

and $\frac{2 b^2}{a} = 12$

$\Rightarrow b^2 = 6a$

Now, using the relation

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

$\Rightarrow 6a = 16 - a^2$

$\Rightarrow a^2 + 6a - 16 = 0$

$\Rightarrow \left( a - 2 \right)\left( a + 8 \right) = 0$

$\Rightarrow a = 2, or - 8$

$b^2 = 12 \text { or }- 48$

Since negative value is not possible, its value is 12.
Thus, the equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{12} = 1$.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Exercise 27.1 | Q 11.07 | Page 14

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