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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

 
Answers (1)
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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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