Q

# Can someone explain - Co-ordinate geometry - JEE Main-9

If a directrix of a hyperbola centered at the origin and passing

through the point $(4,-2\sqrt3)$ is $5x=4\sqrt5$ and its

eccentricity is e , then :

• Option 1)

$4e^{4}-24e^{2}+27=0$

• Option 2)

$4e^{4}-12e^{2}-27=0$

• Option 3)

$4e^{4}-24e^{2}+35=0$

• Option 4)

$4e^{4}+8e^{2}-35=0$

Views

Std eqn of Hyperbola

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

it passes through $(4,-2\sqrt3)$

So,

$\frac{16}{a^{2}}-\frac{12}{b^{2}}=1$

=> $16-12\frac{a^{2}}{b^{2}}=a^{2}$..........................(1)

given eqn of directrix

$5x=4\sqrt5$

=> $x=\frac{4}{\sqrt5}$

Also,

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

=> $\frac{4}{\sqrt5}=\frac{a}{e}$...................................(2)

And we know that  $b^{2}=a^{2}(e^{2}-1)$

$\frac{b^{2}}{a^{2}}=e^{2}-1$................................(3)

from (1),(2) and (3)

=> $16e^{2}-16-12=16\frac{e^{2}}{5}(e^{2}-1)$

=>$4e^{4}-24e^{2}+35=0$

correct option (3)

Option 1)

$4e^{4}-24e^{2}+27=0$

Option 2)

$4e^{4}-12e^{2}-27=0$

Option 3)

$4e^{4}-24e^{2}+35=0$

Option 4)

$4e^{4}+8e^{2}-35=0$

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