Conic Sections Question 58
Question: The line $ x\cos \alpha +y\sin \alpha =p $ will touch the parabola $ y^{2}=4a(x+a) $ , if
Options:
A) $ p\cos \alpha +a=0 $
B) $ p\cos \alpha -a=0 $
C) $ a\cos \alpha +p=0 $
D) $ a\cos \alpha -p=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ x\cos \alpha +y\sin \alpha -p=0 $
…….(i) $ 2ax-yy_1+2a(x_1+2a)=0 $ …….(ii) From (i) and (ii), $ \frac{\cos \alpha }{2a}=\frac{\sin \alpha }{-y}=\frac{-p}{2a(x+2a)} $
therefore $ y=-2\tan \alpha $ and $ x=-p\sec \alpha -2a $
$ \therefore y^{2}=4a(x+a) $
therefore $ 4a^{2}{{\tan }^{2}}\alpha =-4a(p\sec \alpha +a) $
therefore $ p\cos \alpha +a=0 $ .
BETA
