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  • A tangent to the ellipse meets the ellipse <img alt="\mathrm{x^2

A tangent to the ellipse \mathrm{x^2+4 y^2=4} meets the ellipse \mathrm{x^2+2 y^2=6} at P and Q. The tangents at P and Q to the ellipse \mathrm{x^2+2 y^2=6} are making an angle:

Option: 1

\mathrm{\frac{\pi}{3}}


Option: 2

\mathrm{\frac{\pi}{2}}


Option: 3

\mathrm{\frac{2\pi}{3}}


Option: 4

\mathrm{\frac{\pi}{4}}


Answers (1)

best_answer

Let (h, k) be the point of intersection of the tangents to \mathrm{x^2+2 y^2=6} at P and Q. Hence the equation of the line PQ is

\mathrm{ \mathrm{hx}+2 \mathrm{yk}=6 }                                  ......(1)
Tangent at any point \mathrm{(2 \cos \theta, \sin \theta)} to the ellipse \mathrm{x^2+4 y^2=4} is

\mathrm{ \begin{aligned} & 2 \cos \theta x+4 \sin \theta y=4 \\ & \Rightarrow \cos \theta x+2 \sin \theta y=2 \end{aligned} }                          ......(2)

Since (1) and (2) represent the same line,

\mathrm{ \frac{\cos \theta}{\mathrm{h}}=\frac{2 \sin \theta}{2 \mathrm{k}}=\frac{2}{6} }

\mathrm{\Rightarrow \cos \theta=\frac{\mathrm{h}}{3}, \sin \theta=\frac{\mathrm{k}}{3} \Rightarrow \frac{\mathrm{h}^2}{9}+\frac{\mathrm{k}^2}{9}=1}

\mathrm{\Rightarrow h^2+k^2=9=6+3=a^2+b^2}

where the semi - axes of the ellipse \mathrm{x^2+2 y^2=6} are \mathrm{a=\sqrt{6}} and \mathrm{b=\sqrt{3}}

Hence the locus of (h, k) is \mathrm{x^2+y^2=a^2+b^2} which is the director circle of the ellipse

\mathrm{ x^2+2 y^2=6 }
Hence the tangents at P and Q are at right angles.

Posted by

vinayak

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