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Let P(3,3) be a point on the hyperbola, \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1. If the normal to it at P intersects the x-axis at (9,0) and e is its eccentricity then the ordered pair (a^{2},e^{2}) is equal to:
Option: 1 \left (\frac{9}{2},3 \right )
Option: 2 \left (\frac{3}{2},2 \right )
Option: 3 \left (\frac{9}{2},2 \right )
Option: 4 (9,3)

Answers (1)

best_answer

\begin{aligned} &\text { Since, }(3,3) \text { lies on } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\\ &\frac{9}{a^{2}}-\frac{9}{b^{2}}=1\;\;\;\;\ldots(1) \\ \end{aligned}

Now, normal at (3, 3) is

y-3=-\frac{a^{2}}{b^{2}}(x-3)

which passes through (9, 0)

\Rightarrow b^2=2a^2\;\;\;\;\;\;\;\;\;\;\;\dots(2)

\begin{aligned} &\text { So, } \mathrm{e}^{2}=1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=3\\ &\text { Also, } a^{2}=\frac{9}{2}(\text { from }(1) \&\;(2))\\ &\text { Thus, }\left(a^{2}, e^{2}\right)=\left(\frac{9}{2}, 3\right) \end{aligned}

 

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himanshu.meshram

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