Equation of hyperbola in standard form (with transverse axis along x - axis) having length of

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10-2. Parabola, Ellipse, Hyperbola

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The equation of the hyperbola in the standard form (with transverse axis along the $x$ - axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is

A

$\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{18}} = 1$

B

$\frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{27}} = 1$

C

$\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{36}} = 1$

D

$\frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{64}} = 1$

Solution

(c) $\because \frac{{2{b^2}}}{{{a^2}}} = 9$

==> $2{b^2} = 9a$…..$(i)$

Now ${b^2} = {a^2}({e^2} – 1) = \frac{9}{{16}}{a^2}$

==> $a = \frac{4}{3}b$…..$(ii),$ ($e = \frac{5}{4}$)

From $(i)$ and $(ii),$ $b = 6$, $a = 8$

Hence, equation of hyperbola $\frac{{{x^2}}}{{64}} – \frac{{{y^2}}}{{36}} = 1$.

Standard 11

Mathematics

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