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  • The point  lies on the hyperbola <img alt="\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^

The point  P\left ( -2\sqrt{6},\sqrt{3} \right ) lies on the hyperbola \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1 having eccentricity \frac{\sqrt{5}}{2}. If the tangent and normal at P to the hyperbola intesects its conjugate axis at the points Q and R respectively, then QR is equal to:
Option: 1 4\sqrt{3}
Option: 2 6
Option: 3 3\sqrt{6}
Option: 4 6\sqrt{3}

Answers (1)

best_answer


e^{2}= 1+\frac{b^{2}}{a^{2}}\Rightarrow \frac{5}{4}= 1+\frac{b^{2}}{a^{2}}\Rightarrow \frac{b^{2}}{a^{2}}= \frac{1}{4}
and \left ( -2\sqrt{6},\sqrt{3} \right ) lies on it, so

\frac{24}{a^{2}}-\frac{3}{b^{2}}= 1\Rightarrow \frac{24}{a^{2}}-\frac{3\cdot 4}{a^{2}}= 1
\Rightarrow \frac{1}{a^{2}}\left ( 12 \right )= 1\Rightarrow a^{2}= 12
\Rightarrow b^{2}= 3

Equation of tangent at P :

T = 0

\frac{-2\sqrt{6}\cdot x}{12}-\frac{\sqrt{3}y}{3}= 1\Rightarrow slope= \frac{-1}{\sqrt{2}}
For\, Q\, put\: x= 0 \Rightarrow y= -\sqrt{3}
Q\left ( 0,-\sqrt{3} \right )

Normal at p
y-\sqrt{3}= \sqrt{2}\left ( x+2\sqrt{6} \right )
For\, R, \, put\, x= 0\Rightarrow y= 5\sqrt{3}.
R\left ( 0,5\sqrt{3} \right )
\therefore QR= 6\sqrt{3}

Posted by

Kuldeep Maurya

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