Conic Sections Question 63
Question: The point of the contact of the tangent to the parabola $ y^{2}=4ax $ which makes an angle of $ 60^{o} $ with x-axis, is
Options:
A) $ ( \frac{a}{3},\ \frac{2a}{\sqrt{3}} ) $
B) $ ( \frac{2a}{\sqrt{3}},\ \frac{a}{3} ) $
C) $ ( \frac{a}{\sqrt{3}},\ \frac{2a}{3} ) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ m=\tan \theta =\tan 60{}^\circ =\sqrt{3} $
The equation of tangent at $ (h,k) $ to $ y^{2}=4ax $ is $ yk=2a(x+h) $
Comparing, we get $ m=\sqrt{3}=\frac{2a}{k} $ or $ k=\frac{2a}{\sqrt{3}} $ and $ h=\frac{a}{3} $ .
BETA
