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If  \mathrm {5 x+9=0 }  is the directrix of the hyperbola   \mathrm {16 x^2-9 y^2=144 }  , then its corresponding focus is
 

Option: 1

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


Option: 2

(-5,0)


Option: 3

(5,0)


Option: 4

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


Answers (1)

best_answer

The given hyperbola is  \mathrm {\frac{16 x^2}{144}-\frac{9 y^2}{144}=1} 

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

Now, eccentricity \mathrm { (e)=\sqrt{1+\frac{16}{9}}=\frac{5}{3} } 
So, directrix of the hyperbola are  \mathrm { x= \pm \frac{a}{e}= \pm \frac{9}{5} } 
It is given that the directrix of the hyperbola is  \mathrm { x=-\frac{9}{5}. }


\mathrm { \therefore \quad }  Required corresponding focus is  \mathrm { (-a e, 0)=(-5,0). } 

Posted by

Irshad Anwar

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