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Given the equation of a hyperbola \(H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and its directrix is \(x=\sqrt{\frac{10}{81}}\) with a focus at \((\sqrt{10}, 0,\) then find the value of \(9\left(e+I^{2}\right),\) where I is length of latus rectum is

(1) 2697

(2) 2597

(3) 2487

(4) 2587

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1 Answer

+1 vote
by (51.7k points)

Correct option is: (4) 2587 

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

Directrix : \(x=\frac{\sqrt{10}}{9}=\frac{a}{e}\)  .....(i)

Focus: \((a e, 0) \equiv a e=\sqrt{10}\)    .....(ii)

\((i) \times (ii)\)

\(a^{2}=\frac{10}{9}\)

Dividing equation (ii) by (i) we get,

\(e^{2}=9 \Rightarrow e=3\)

Also, \(e^{2}=9=1+\frac{9 b^{2}}{10}\)

\(\Rightarrow b^{2}=\frac{80}{9}\)

\(I=2 \times \frac{b^{2}}{a}=\frac{2 \times \frac{80}{9}}{\frac{\sqrt{10}}{3}}=\frac{160}{3 \sqrt{10}}\)

Now, \(9\left(e+l^{2}\right)=9\left[3+\frac{25600}{9 \times 10}\right]\)

\(9\left(3+\frac{160}{3 \sqrt{10}}\right)=27+2560\)

= 2587

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