Applications Of Derivatives Question 37
Question: Find the angle between the tangent to the curve $ y^{2}=2ax $ at the points where x = a/2.
Options:
A) $ 180{}^\circ $
B) $ 90{}^\circ $
C) $ 0{}^\circ $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] We have, $ y^{2}=2ax $ Put $ x=\frac{a}{2};y^{2}=2a( \frac{a}{2} )\Rightarrow y=\pm a $
$ \therefore $ The points are $ ( \frac{a}{2},a ) $ and $ ( \frac{a}{2},-a ) $ Differentiating (i) with respect to x, we get $ 2y\frac{dy}{dx}=2a\Rightarrow \frac{dy}{dx}=\frac{a}{y} $ At $ ( \frac{a}{2},a );\frac{dy}{dx}=\frac{a}{y}=\frac{a}{a}=1=m_1,(say) $ At $ ( \frac{a}{2},-a );\frac{dy}{dx}=\frac{a}{y}=\frac{a}{-a}=-1=m_2(say) $ Since $ m_1m_2=-1 $ , the two tangents are at right angles.
BETA
