Conic Sections Question 93
Question: The equation of the tangent to the parabola $ y^{2}=16x $ , which is perpendicular to the line $ y=3x+7 $ is
[MP PET 1998]
Options:
A) $ y-3x+4=0 $
B) $ 3y-x+36=0 $
C) $ 3y+x-36=0 $
D) $ 3y+x+36=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
Line perpendicular to given line, $ 3y+x=\lambda $
$ y=\frac{-1}{3}x+\frac{\lambda }{3} $ . Here, $ m=\frac{-1}{3},c=\frac{\lambda }{3} $
If we compare $ y^{2}=16x $ with $ y^{2}=4ax $ then $ a=4 $ , Condition for tangency is, $ c=\frac{a}{m}\Rightarrow \frac{\lambda }{3}=\frac{4}{(-1/3)}\Rightarrow \lambda =-36 $
Required equation is; $ x+3y+36=0 $ .
BETA
