JEE PYQ: Ellipse Question 24
Question 24 - 2019 (11 Jan Shift 1)
If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ 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:
(a) $\frac{1}{4x^2} + \frac{1}{2y^2} = 1$ (b) $\frac{x^2}{4} + \frac{y^2}{2} = 1$ (c) $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$ (d) $\frac{x^2}{2} + \frac{y^2}{4} = 1$
Show Answer
Answer: (c) $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$
Solution
Ellipse: $\frac{x^2}{2} + y^2 = 1$
Tangent: $\frac{\sqrt{2}\cos\theta \cdot x}{2} + y\sin\theta = 1$
Intercepts: $P\left(\frac{\sqrt{2}}{\cos\theta}, 0\right)$, $Q\left(0, \frac{1}{\sin\theta}\right)$
Midpoint $(h, k)$: $h = \frac{1}{\sqrt{2}\cos\theta}$, $k = \frac{1}{2\sin\theta}$
Locus: $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$
BETA
