Conic Sections Question 142
Question: The equations of the normals at the ends of latus rectum of the parabola $ y^{2}=4ax $ are given by
Options:
A) $ x^{2}-y^{2}-6ax+9a^{2}=0 $
B) $ x^{2}-y^{2}-6ax-6ay+9a^{2}=0 $
C) $ x^{2}-y^{2}-6ay+9a^{2}=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The co-ordinates of the ends of the latus rectum of the parabola $ y^{2}=4ax $ are $ (a,2a) $ and $ (a,-2a) $ respectively. The equation of the normal at $ (a,2a) $ to $ y^{2}=4ax $ is $ y-2a=-\frac{2a}{2a}(x-a) $
$ [ Usingy-{y_1}=-\frac{y_1}{2a}(x-x_1) ] $
or $ x+y-3a=0 $ ……(i) Similarly, the equation of the normal at $ (a,-2a) $ is $ x-y-3a=0 $ …… (ii) The combined equation of (i) and (ii) is $ x^{2}-y^{2}-6ax+9a^{2}=0 $ .
BETA
