Conic Sections Question 414
Question: Which one of the following curves cuts the parabola $ y^{2}=4ax $ at right angles
[IIT 1994]
Options:
A) $ x^{2}+y^{2}=a^{2} $
B) $ y={e^{-x/2a}} $
C) $ y=ax $
D) $ x^{2}=4ay $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y^{2}=4ax $
therefore $ 2y{{( \frac{dy}{dx} )}_1}=4a $
therefore $ {{( \frac{dy}{dx} )}_1}=\frac{2a}{y} $ …..(i) Taking curve $ y={e^{-x/2a}} $
$ {{( \frac{dy}{dx} )}_2}={e^{-x/2a}}( -\frac{1}{2a} ) $
$ =-\frac{y}{2a} $
…..(ii) Both curves cut orthogonally if, $ {{( \frac{dy}{dx} )}_1}{{( \frac{dy}{dx} )}_2}=-1 $
therefore $ ( -\frac{y}{2a} ).( \frac{2a}{y} )=-1 $ .
BETA
