Question

The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse $x^2+2y^2=2$ between the coordinate axes is:

• A. $\frac{1}{x^2}+\frac{1}{2y^2}=1$
• B. $\frac{1}{{4x}^2}+\frac{1}{2y^2}=1$
• C. $\frac{1}{{2x}^2}+\frac{1}{4y^2}=1$
• D. $\frac{1}{{2x}^2}+\frac{1}{y^2}=1$

Answer 1

The correct option is C $\frac{1}{{2x}^2}+\frac{1}{4y^2}=1$

Let the point of contact be $R\equiv (\sqrt{2}\cos \theta ,\sin \theta )$

equation of tangent $AB$ is $\frac{x}{\sqrt{2}}\cos \theta +y\sin \theta =1$

$\Rightarrow A\equiv (\sqrt{2}\sec \theta ,0);B\equiv (0,\csc \theta )$

Let the middle point $Q$ of $AB$ be $(h,k)\Rightarrow h=\frac{\sec \theta }{\sqrt{2}},k=\frac{\csc \theta }{2}$

$\Rightarrow \cos \theta =\frac{1}{h\sqrt{2}},\sin \theta =\frac{1}{2k}\Rightarrow \frac{1}{2h^2}+\frac{1}{4k^2}=1$

$\therefore$ Required locus is $\frac{1}{2x^2}+\frac{1}{4y^2}=1.$

Trick: The locus of mid-points of the portion of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ intercepted between axes is $a^2{y}^2+b^2{x}^2=4x^2{y}^2$

i.e., $\frac{a^2}{4x^2}+\frac{b^2}{4y^2}=1$ or $\frac{1}{2x^2}+\frac{1}{4y^2}=1.$