Question

If the tangents are drawn to the ellipse $x^2+2y^2=2$, then the locus of the mid-point of the intercept made by the tangents between the coordinate-axes is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$x^2+2y^2=2\Rightarrow \frac{x^2}{(\sqrt{2}{)}^2}+y^2=1\text{Equation of tangent in parametric form is}\frac{x^2}{\sqrt{2}}cos\theta +ysin\theta =1\text{Intersects x-axis at A}(\frac{\sqrt{2}}{cos\theta },\theta ),\text{y-axis at B}(0,\frac{1}{sin\theta })\text{Let p(h, k) be the mid-point of intercept AB}\therefore h=\frac{\sqrt{2}cos\theta }{2},k=\frac{1}{2sin\theta }\Rightarrow cos\theta =\frac{1}{\sqrt{2}h},sin\theta =\frac{1}{2k}\therefore \frac{1}{2h^2}+\frac{1}{4k^2}=1\therefore \text{(locus of h, k) is}\frac{1}{2x^2}+\frac{1}{4y^2}=1$