Conic Sections Question 131
Question: The equation of normal to the parabola at the point $ ( \frac{a}{m^{2}},\ \frac{2a}{m} ) $ ,is
[RPET 1987]
Options:
A) $ y=m^{2}x-2mx-am^{3} $
B) $ m^{3}y=m^{2}x-2am^{2}-a $
C) $ m^{3}y=2am^{2}-m^{2}x+a $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ y-\frac{2a}{m}=-\frac{2a/m}{2a}( x-\frac{a}{m^{2}} ) $
therefore $ y-\frac{2a}{m}=\frac{-1}{m}( x-\frac{a}{m^{2}} ) $
therefore $ m^{3}y+m^{2}x-2am^{2}-a=0 $ .
BETA
