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  • From a point, perpendicular tangents are drawn to the ellipse<img alt="\mathrm{4 x^2+9 y^2=36}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B4%20x%5E2&plus;9%20y%5E2%3D36%7D

From a point, perpendicular tangents are drawn to the ellipse\mathrm{4 x^2+9 y^2=36}. The chord of contact touches a circle concentric with the given ellipse. If the sum of the maximum and minimum areas of the circles is \mathrm{\pi k \text {. Then } 52 \times 117 k=}

Option: 1

1261


Option: 2

1361


Option: 3

1300


Option: 4

1200


Answers (1)

best_answer

\mathrm{\text { Ellipse: } \frac{x^2}{9}+\frac{y^2}{4}=1}

Perpendicular tangents intersect on director circle. Equation of director circle is \mathrm{x^2+y^2=13}

\mathrm{\text { Any point on director circle is } P(\sqrt{13} \cos \theta, \sqrt{13} \sin \theta)}

Equation of chord of contact of ellipse w.r.t. point P is

\mathrm{(4 \sqrt{13} \cos \theta) x+(9 \sqrt{13} \sin \theta) y-36=0}

\mathrm{\text { Let it touches circle } x^2+y^2=r^2}

\mathrm{\begin{aligned} & \therefore r=\left|\frac{-36}{\sqrt{208 \cos ^2 \theta+1053 \sin ^2 \theta}}\right| \\ & \therefore r^2=\frac{36}{208+845 \sin ^2 \theta} \\ & \therefore r_{\text {max }}^2=\frac{36}{208+0}=\frac{9}{52} \text { and } r_{\text {min }}^2=\frac{36}{1053}=\frac{4}{117} \end{aligned}}

Max. area of circle + Min. area of circle

\mathrm{=\pi\left(\frac{9}{52}\right)+\pi\left(\frac{4}{117}\right)=\frac{1261 \pi}{52 \times 117}}

Posted by

Ajit Kumar Dubey

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