Conic Sections Question 151
Question: Equation of any normal to the parabola $ y^{2}=4a(x-a) $ is
Options:
A) $ y=mx-2am-am^{3} $
B) $ y=m(x+a)-2am-am^{3} $
C) $ y=m(x-a)+\frac{a}{m} $
D) $ y=m(x-a)-2am-am^{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let normal at $ (h,k) $ be $ y=mx+c $
then, $ k=mh+c $ also $ k^{2}=4a(h-a) $
slope of tangent at $ (h,k) $ is $ m_1 $ then on differentiating equation of parabola. $ 2ym_1=4a $
therefore $ m_1=\frac{2a}{k} $ also $ mm_1=-1 $
therefore $ m=-\frac{k}{2a}, $ solving and replacing $ h,k) $ by $ (x,y) $
therefore $ y=m(x-a)-2am-am^{3} $ .
BETA
