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If chord of contact of the tangents drawn from the point  (\alpha, \beta)  to the ellipse   \frac{\mathrm x^2}{\mathrm a^2}+\frac{\mathrm y^2}{\mathrm b^2}=1,touches the circle \mathrm x^2+\mathrm y^2=\mathrm c^2, then the locus of the point (\alpha, \beta)  is

Option: 1

\frac{\mathrm x^2}{\mathrm a^2}+\frac{\mathrm y^2}{\mathrm b^2}=\frac{1}{\mathrm c^2}


Option: 2

\frac{\mathrm x^2}{\mathrm a^2}+\frac{\mathrm y^2}{\mathrm b^2}=\frac{1}{\mathrm c^4}


Option: 3

\frac{\mathrm x^2}{\mathrm a^4}+\frac{\mathrm y^2}{\mathrm b^4}=\frac{1}{\mathrm c^2}


Option: 4

 None of these


Answers (1)

best_answer

\frac{\mathrm x \alpha}{\mathrm a^2}+\frac{\mathrm y \beta}{\mathrm b^2}=1  touches the circles  \mathrm x^2+\mathrm y^2=\mathrm c^2


\therefore \quad  perpendicular distance from centre of the circle to the line = radius of the circle
i.e., \frac{|-1|}{\sqrt{\frac{\alpha^2}{a^4}+\frac{\beta^2}{b^4}}}=c
Locus of  (\alpha, \beta)  is \mathrm c^2\left(\frac{\mathrm x^2}{\mathrm a^4}+\frac{\mathrm y^2}{\mathrm b^4}\right)=1 \quad i.e., \frac{\mathrm x^2}{a^4}+\frac{\mathrm y^2}{\mathrm b^4}=\frac{1}{\mathrm c^2}

Hence (3) is the correct answer.

Posted by

sudhir.kumar

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