Question

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. $\frac{a^2}{x^2}+\frac{b^2}{y^2}=3$
• D. $\frac{a^2}{x^2}+\frac{b^2}{y^2}=4$

Answer 1

The correct option is D $\frac{a^2}{x^2}+\frac{b^2}{y^2}=4$

Let mid-point of part $PQ$ which is in between the axis is $R(x_1,y_1)$, then coordinates of $P$ and $Q$ will be $(2x_1,0)$ and $(0,2y_1)$, respectively.
$\therefore$ Equation of line $PQ$ is $\frac{x}{2x_1}+\frac{y}{2y_1}=1$
$\Rightarrow y=-\left(\frac{y_1}{x_1}\right)x+2y_1$
If this line touches the ellipse
$\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$
then it will satisfy the condition,
$c^2=a^2{m}^2+b^2$
So, $(2y_1{)}^2=a^2{\left(\frac{-y_1}{x_1}\right)}^2+b^2$
$\Rightarrow 4y_1^2=\left\{\frac{a^2{y}_1^2}{x_1^2}\right\}+b^2$
$\Rightarrow 4=\left(\frac{a^2}{x_1^2}\right)+\left(\frac{b^2}{y_1^2}\right)\Rightarrow \left(\frac{a^2}{x_1^2}\right)+\left(\frac{b^2}{y_1^2}\right)=4$
$\therefore$ Required locus of $(x_1,y_1)$ is
$\frac{a^2}{x^2}+\frac{b^2}{y^2}=4$