EllipseHardBloom L3

Question

If $P(\theta)$ and $Q\left( \frac{\pi}{2} + \theta \right)$ are two points on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, then locus of midpoint of $PQ$ is:

Options

A.$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{1}{2}$
B.$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2$
C.$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 4$
D.$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 8$

Solution

Sol.(a) $2h = a(cos\theta - sin\theta)$

$$2k = b(cos\theta + sin\theta)$$

( $- asin\theta,bcos\theta$ )

$$\begin{matrix} & \ \Rightarrow \left( \frac{2h}{a} \right)^{2} + \left( \frac{2k}{b} \right)^{2} = 2\left( \cos^{2}\theta + \sin^{2}\theta \right) \\ & \ \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{1}{2} \end{matrix}$$

Create a free account to view solution

View Solution Free
Topic: Ellipse·Practice all Ellipse questions

More Ellipse Questions