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  • The area of the quadrilateral formed by the common tangents of the circle <img alt="\mathrm{x^2+y^2=c^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2%3Dc%5

The area of the quadrilateral formed by the common tangents of the circle \mathrm{x^2+y^2=c^2} and the ellipse \mathrm{ {\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}, b<c<a~ is~\frac{2 c^2 k}{\sqrt{\left(a^2-c^2\right)\left(c^2-b^2\right)}}}, where \mathrm{k}=

Option: 1

\mathrm{ a^2+b^2}


Option: 2

\mathrm{ \sqrt{a^2+b^2}}


Option: 3

\mathrm{ a^2-b^2}


Option: 4

\mathrm{ \sqrt{a^2-b^2}}


Answers (1)

best_answer

Let m be the slope of the common tangent A B of the given circle and the ellipse in the first quadrant, intersecting the x and y-axes at A and B respectively.

Then c \mathrm{\sqrt{1+m^2}=\sqrt{a^2 m^2+b^2} }

\mathrm{\Rightarrow \quad m=-\sqrt{\frac{c^2-b^2}{a^2-c^2}}}

Thus equation of the line passing through A and B is

\begin{gathered} \mathrm{y=m x+c \sqrt{1+m^2}} \\ \mathrm{A=\left(c \sqrt{\frac{a^2-b^2}{c^2-b^2}}, 0\right), B=\left(0, c \sqrt{\frac{a^2-b^2}{a^2-c^2}}\right) }\end{gathered}
Hence

Due to symmetry, the quadrilateral formed by the common tangents will be a rhombus having diagonals along the axes of coordinates of lengths \mathrm{ 2 c \sqrt{\frac{a^2-b^2}{c^2-b^2}} ~and ~2 c \sqrt{\frac{a^2-b^2}{a^2-c^2}}}

Required area = \mathrm{\frac{1}{2} \times 2 c \sqrt{\frac{a^2-b^2}{c^2-b^2}} \times 2 c \sqrt{\frac{a^2-b^2}{a^2-c^2}}=\frac{2 c^2\left(a^2-b^2\right)}{\sqrt{\left(a^2-c^2\right)\left(c^2-b^2\right)}}}
 

Posted by

Ritika Jonwal

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