The normal to the hyperbola
${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$ at the point $\left( {8,3\sqrt 3 } \right)$ on it passes through the point :
Solution
<p>Given hyperbola : ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$</p>
<p>$\because$ It passes through $(8,3\sqrt 3 )$</p>
<p>$\because$ ${{64} \over {{a^2}}} - {{27} \over 9} = 1 \Rightarrow {a^2} = 16$</p>
<p>Now, equation of normal to hyperbola</p>
<p>${{16x} \over 8} + {{9y} \over {3\sqrt 3 }} = 16 + 9$</p>
<p>$\Rightarrow 2x + \sqrt 3 y = 25$ ...... (i)</p>
<p>$\left( { - 1,9\sqrt 3 } \right)$ satisfies (i)</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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