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

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

Option: 1

2


Option: 2

\frac{1}{2}


Option: 3

1


Option: 4

\sqrt{2}


Answers (1)

best_answer

Let m be the slope of the common tangent A B of the given circle and ellipse in the first quadrant intersecting x and y axes at A and B respectively. Then \mathrm{\mathrm{c} \sqrt{1+\mathrm{m}^2}=\sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2} }
\mathrm{\Rightarrow \quad \mathrm{m}=-\sqrt{\frac{\mathrm{c}^2-\mathrm{b}^2}{\mathrm{a}^2-\mathrm{c}^2}} }
Thus equation of the line passing through\mathrm{ \mathrm{A} \, \, and\, \, B is y=m x+c \sqrt{1+m^2}. } Hence


A\mathrm{ \equiv\left(c \sqrt{\frac{a^2-b^2}{c^2-b^2}}, 0\right), B \equiv\left(0, c \sqrt{\frac{a^2-b^2}{a^2-c^2}}\right) }
Due to symmetry the quadrilateral formed by the common tangents will be a rhom having diagonals along axes of coordiantes of length
\mathrm{ 2 c \sqrt{\frac{a^2-b^2}{c^2-b^2}} \text { and } 2 c \sqrt{\frac{a^2-b^2}{a^2-c^2}} }
Hence 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{a^2-c^2} \sqrt{c^2-b^2}} }
 

Posted by

Sanket Gandhi

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