Gujarati
10-2. Parabola, Ellipse, Hyperbola
medium

If the latus rectum of an hyperbola be 8 and eccentricity be $3/\sqrt 5 $, then the equation of the hyperbola is

A

$4{x^2} - 5{y^2} = 100$

B

$5{x^2} - 4{y^2} = 100$

C

$4{x^2} + 5{y^2} = 100$

D

$5{x^2} + 4{y^2} = 100$

Solution

(a) $\frac{{2{b^2}}}{a} = 8$ and $\frac{3}{{\sqrt 5 }} = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} $ or $\frac{4}{5} = \frac{{{b^2}}}{{{a^2}}}$

==> $a = 5$, $b = 2\sqrt 5 $.

Hence the required equation of hyperbola is $\frac{{{x^2}}}{{25}} – \frac{{{y^2}}}{{20}} = 1$

==> $4{x^2} – 5{y^2} = 100$.

Standard 11
Mathematics

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