Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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# Find the Equation of the Hyperbola Whose Vertices Are at (± 6, 0) and One of the Directrices Is X = 4. - Mathematics

Answer in Brief

Find the equation of the hyperbola whose vertices are at (± 6, 0) and one of the directrices is x = 4.

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#### Solution

The Vertices of the hyperbola are $\left( \pm 6, 0 \right)$

∴ $a = 6$

⇒ a2 = 36
Now, x = 4

$\frac{a}{e} = 4$

$\Rightarrow e = \frac{3}{2} \left[ \because a = 6 \right]$

Now,

$\left( ae \right)^2 = a^2 + b^2$

$\Rightarrow \left( 6 \times \frac{3}{2} \right)^2 = 6^2 + b^2$

$\Rightarrow 81 - 36 = b^2$

$\Rightarrow b^2 = 45$

Therefore, the equation of the hyperbola is $\frac{x^2}{36} - \frac{y^2}{45} = 1$.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Exercise 27.1 | Q 7.5 | Page 14
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