JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Ellipse

Write the ellipse as $\frac{x^2}{2}+y^2=1$, so $a^2=2$ and $b^2=1$, and take the tangent at parameter $\theta$: $\frac{x\cos \theta }{\sqrt{2}}+y\sin \theta =1$. Its intercepts are $\left(\frac{\sqrt{2}}{\cos \theta },0\right)$ and $\left(0,\frac{1}{\sin \theta }\right)$, so the midpoint $(h,k)$ satisfies $2h=\frac{\sqrt{2}}{\cos \theta }$ and $2k=\frac{1}{\sin \theta }$, giving $\cos \theta =\frac{\sqrt{2}}{2h}$ and $\sin \theta =\frac{1}{2k}$. Imposing ${\cos }^2\theta +{\sin }^2\theta =1$ gives $\frac{2}{4h^2}+\frac{1}{4k^2}=1$, that is $\frac{1}{2h^2}+\frac{1}{4k^2}=1$; the midpoints lie on $\frac{1}{2x^2}+\frac{1}{4y^2}=1$.