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  • Confused! kindly explain, - Co-ordinate geometry - JEE Main-4

If 5x+9=0 is the directrix of the hyperbola 16x^{2}-9y^{2}=144,

then its corresponding focus is : 

  • Option 1)

    (5,0)

  • Option 2)

    (\frac{-5}{3},0)

  • Option 3)

    (\frac{5}{3},0)

  • Option 4)

    (-5 , 0)

Answers (1)

eqn of hyperbola 16x^{2}-9y^{2}=144,

\frac{x^{2}}{\frac{144}{16}}-\frac{y^{2}}{\frac{144}{9}}=1

\frac{x^{2}}{9}-\frac{y^{2}}{16}=1

Directrix of hyperbola 5x+9=0\Rightarrow x=-\frac{9}{5}

Now, from the eqn of hyperbola

a^{2}=9 , b^{2}=16

As we know eccentricity = \frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{5}{3}

Required focus is (-ae,0) =(-3\cdot \frac{5}{3},0)

                                                =(-5,0)

Option (4) is correct answer.

 


Option 1)

(5,0)

Option 2)

(\frac{-5}{3},0)

Option 3)

(\frac{5}{3},0)

Option 4)

(-5 , 0)
Posted by

Vakul

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