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  • If tangents are drawn to the ellipse , then the locus of the midp

If tangents are drawn to the ellipse \mathrm{x^2+2 y^2=2}, then the locus of the midpoint of the intercept made by the tangents between the coordinate axes is

Option: 1

\mathrm{\frac{1}{2 x^2}+\frac{1}{4 y^2}=1}


Option: 2

\mathrm{\frac{1}{4 x^2}+\frac{1}{2 y^2}=1}


Option: 3

\mathrm{\frac{x^2}{2}+\frac{y^2}{4}=1}


Option: 4

\mathrm{\frac{x^2}{4}+\frac{y^2}{2}=1}


Answers (1)

best_answer

Let the point of contact be \mathrm{R \equiv(\sqrt{2} \cos \theta, \sin \theta)}. Equation of tangent AB is

\mathrm{\begin{aligned} & \frac{x}{\sqrt{2}} \cos \theta+y \sin \theta=1 \\ & \Rightarrow \quad A \equiv(\sqrt{2} \sec \theta, 0), B \equiv(0, \operatorname{cosec} \theta) \end{aligned}}

Let the mid-point Q of AB  be (h, k)

\mathrm{\begin{aligned} & \Rightarrow h=\frac{\sec \theta}{\sqrt{2}}, k=\frac{\operatorname{cosec} \theta}{2} \\ & \Rightarrow \cos \theta=\frac{1}{h \sqrt{2}}, \sin \theta=\frac{1}{2 k} \\ & \Rightarrow \frac{1}{2 h^2}+\frac{1}{4 k^2}=1 \end{aligned}}

\mathrm{\text { Thus, the required locus is } \frac{1}{2 x^2}+\frac{1}{4 y^2}=1}

Posted by

shivangi.bhatnagar

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