Eccentricity of hyperbola 5{x^2} - 4{y^2} + 20x + 8y = 4 is

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10-2. Parabola, Ellipse, Hyperbola

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The eccentricity of the hyperbola $5{x^2} - 4{y^2} + 20x + 8y = 4$ is

A

$\sqrt 2 $

B

$\frac{3}{2}$

C

$2$

D

$3$

Solution

(b) Given equation of hyperbola is $5{x^2} – 4{y^2} + 20x + 8y = 4$

$5{(x + 2)^2} – 4\,{(y – 1)^2} = 20$ ==> $\frac{{{{(x + 2)}^2}}}{4} – \frac{{{{(y – 1)}^2}}}{5} = 1$

From ${b^2} = {a^2}({e^2} – 1)$, $5 = 4({e^2} – 1)$

$ \Rightarrow {e^2} = 9/4 $

$\Rightarrow e = 3/2$.

Standard 11

Mathematics

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(JEE MAIN-2019)