Eccentricity of hyperbola conjugate to hyperbola frac{{{x^2}}}{4}

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10-2. Parabola, Ellipse, Hyperbola

normal

Eccentricity of the hyperbola conjugate to the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1$ is

$\frac{2}{{\sqrt 3 }}$

$2$

$\sqrt 3 $

$\frac{4}{3}$

Solution

$e_1^2 = 1 + \frac{{{b^2}}}{{{a^2}}}$ $=$ $1 +$$\frac{{12}}{4}$ $= 4$ $ \Rightarrow$ $ e_1 = 2$ ;
now $\frac{1}{{e_1^2}} + \frac{1}{{e_2^2}}$ $= 1$
$\frac{1}{{e_2^2}}$ $=$ $ 1 -$ $\frac{1}{4}$ $ =$ $\frac{3}{4}$
$ \Rightarrow$ $e_2^2$ $=$ $\frac{4}{3}$ $\Rightarrow$ $e_2 $ $= $ $\frac{2}{{\sqrt 3 }}$

Standard 11

Mathematics

Let $\mathrm{P}$ be a point on the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$, in the first quadrant such that the area of triangle formed by $\mathrm{P}$ and the two foci of $\mathrm{H}$ is $2 \sqrt{13}$. Then, the square of the distance of $\mathrm{P}$ from the origin is

hard

(JEE MAIN-2024)

The eccentricity of the hyperbola $\frac{{{x^2}}}{{16}} – \frac{{{y^2}}}{{25}} = 1$ is

easy

The length of transverse axis of the parabola $3{x^2} – 4{y^2} = 32$ is

easy

For the hyperbola $\frac{{{x^2}}}{9} – \frac{{{y^2}}}{3} = 1$ the incorrect statement is :