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  • The locus of the point of intersection of the perpendicular tangents to the ellipse 

The locus of the point of intersection of the perpendicular tangents to the ellipse \mathrm{\frac{x^2}{9}+\frac{y^2}{4}=1} is
 

Option: 1

\mathrm{x^2+y^2=9}


Option: 2

\mathrm{x^2+y^2=4}


Option: 3

\mathrm{x^2+y^2=13}


Option: 4

\mathrm{x^2+y^2=5}


Answers (1)

best_answer

The locus of point of intersection of two perpendicular tangents drawn on the ellipse is \mathrm{x^2+y^2=a^2+b^2} which is called “director circle”.
Given ellipse is \mathrm{\frac{x^2}{9}+\frac{y^2}{4}=1}
\mathrm{\therefore \text { Locus is } x^2+y^2=9+4 \text {, i.e. } x^2+y^2=13}

Posted by

Kuldeep Maurya

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