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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%

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 \mathrm{\frac{k c^2\left(a^2-b^2\right)}{\sqrt{\left(a^2-c^2\right)\left(c^2-b^2\right)}}}., where \mathrm{k=}

Option: 1

1


Option: 2

2


Option: 3

\frac{1}{2}


Option: 4

3


Answers (1)

best_answer

Let m be the slope of the common tangent AB 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 m=-\sqrt{\frac{c^2-b^2}{a^2-c^2}} }

Thus equation of the line passing through A and B is \mathrm{y=m x+c \sqrt{1+m^2}}. Hence

\mathrm{ A \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 rhombus having

diagonals along axes of coordinates of length \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}}.}

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

Ritika Kankaria

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