Question bank/JEE Main/Mathematics

Tangents are drawn to the ellipse $x^{2} + 2y^{2} = 2$. The
locus of the midpoint of the intercept made by tangents between the axes
is $\frac{1}{2x^{2}} + \frac{1}{ky^{2}} = 1$ then 4k is
fcUnqvksa dk fcUnqiFk
$\frac{1}{2x^{2}} + \frac{1}{ky^{2}} = 1$gS rc 4k

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[IMAGE] Consider any point $P(\sqrt{2}cos\theta,sin\theta)$
Tangent at this point is $\sqrt{2}cos\theta x + 2sin\theta y = 2$
Now it intersect at x-axis at $(\sqrt{2}sec\theta,0)$ and y-axis at
$(0,cosec\theta)$
Now find its midpoint and eliminate $\theta$ by squaring and adding.
[IMAGE] ekuk fd fcUnqP(√2 cos 𝜃, sin 𝜃)
bl fcUnq ij √2 cos 𝜃 𝑥 + 2 sin 𝜃 𝑦 = 2
vc
vc bl fcUnq Kkr dhft, rFkk𝜃

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx