Conic-Sections Question 574
Question: If $ P(\theta ) $ and $ Q( \frac{\pi }{2}+\theta ) $ are two points on the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ , then locus of the mid-point 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}}=4 $
C) $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Clearly P is $ (acos\theta ,bsin\theta ) $ and Q is $ (-asin\theta ,b\sin \theta ) $ so the midpoint (h, k) or PQ will be given by $ h=\frac{a\cos \theta -a\sin \theta }{2} $ And $ k=\frac{b\sin \theta +b\cos \theta }{2} $
$ \therefore \frac{4h^{2}}{a^{2}}+\frac{4k^{2}}{b^{2}}=2\Rightarrow \frac{h^{2}}{a^{2}}+\frac{k^{2}}{b^{2}}=\frac{1}{2} $
BETA
