Q69 / 120 · VITEEE 2013 · Mathematics
The locus of mid points of tangents intercepted between the axes of ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ will be
- A$\frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=1$
- B$\frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=2$
- C$\frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=3$
- Correct answerD$\frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4$
Solution
Let mid-point of part PQ which is in between the axis is $\mathrm{R}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$, then coordinates of $P$ and $Q$ will be $\left(2 x_{1}, 0\right)$ and $\left(0,2 \mathrm{y}_{1}\right)$, respectively.
$\therefore$ Equation of line $\mathrm{PQ}$ is $\frac{\mathrm{x}}{2 \mathrm{x}_{1}}+\frac{\mathrm{y}}{2 \mathrm{y}_{1}}=1$
$\Rightarrow y=-\left(\frac{y_{1}}{x_{1}}\right) x+2 y_{1}$
If this line touches the ellipse
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
then it will satisfy the condition, $c^{2}=a^{2} m^{2}+b^{2}$
So, $\left(2 y_{1}\right)^{2}=a^{2}\left(\frac{-y_{1}}{x_{1}}\right)^{2}+b^{2}$
$\begin{array}{l}
\Rightarrow 4 y_{1}^{2}=\left\{\frac{a^{2} y_{1}^{2}}{x_{1}^{2}}\right\}+b^{2} \\
\Rightarrow 4=\left(\frac{a^{2}}{x_{1}^{2}}\right)+\left(\frac{b^{2}}{y_{1}^{2}}\right) \Rightarrow\left(\frac{a^{2}}{x_{1}^{2}}\right)+\left(\frac{b^{2}}{y_{1}^{2}}\right)=4
\end{array}$
$\therefore$ Required locus of $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is
$\left(\frac{\mathrm{a}^{2}}{\mathrm{x}^{2}}+\frac{\mathrm{b}^{2}}{\mathrm{y}^{2}}\right)=4$
Hence, the correct answer is option: 'D'.