Ellipse Question - None

Question

If tangents are drawn to the ellipse x² + 2y² = 2 at all points on the ellipse other than its four vertices, then the mid-points of the tangents intercepted between the coordinate axes lie on the curve

Accepted Answer

Given equation of ellipse is x² + 2y² = 2 , which can be written as $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$
Let P be a point on the ellipse, other than its four vertices. Then, the parametric coordinates of P be ($\sqrt2$ cos$\theta$, sin$\theta$)

Now, the equation of tangent at P is
$\frac{x \sqrt{2} \cos \theta}{2}+\frac{y \sin \theta}{1}=1$ [$\because$ equation of tangent at (x, y) is given by T = 0 $\Rightarrow \frac{x x_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=1$]
$\Rightarrow \quad \frac{x}{\sqrt{2} \sec \theta}+\frac{y}{cosec \theta}=1$
$\because A(\sqrt{2} \sec \theta, 0) \text { and } B(0, cosec \theta)$
Let mid-point of AB be R(h, k), then
$h=\frac{\sqrt{2} \sec \theta}{2} \text { and } k=\frac{cosec \theta}{2}$
$2 h=\sqrt{2} \sec \theta \text { and } 2 k=cosec \theta$
$\Rightarrow \quad \cos \theta=\frac{1}{\sqrt{2} h} \text { and } \sin \theta=\frac{1}{2 k}$
We know that, $\cos ^{2} \theta+\sin ^{2} \theta=1$
$\therefore \quad \frac{1}{2 h^2}+\frac{1}{4 k^{2}}=1$
So, locus of (h, k) is $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$

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