Applications-Of-Derivatives Question 445
Question: The point of the curve $ y^{2}=2(x-3) $ at which the normal is parallel to the line $ y-2x+1=0 $ is
[MP PET 1998]
Options:
A) (5,2)
B) $ ( -\frac{1}{2},-2 ) $
C) (5, ?2)
D) $ ( \frac{3}{2},,2 ) $
Show Answer
Answer:
Correct Answer: C
Solution:
Given $ y^{2}=2(x-3) $ …..(i) Differentiate w.r.t. x, $ 2y.\frac{dy}{dx}=2\Rightarrow \frac{dy}{dx}=\frac{1}{y} $ Slope of the normal $ =\frac{-1}{( \frac{dy}{dx} )}=-y $ Slope of the given line $ =2 $
$ \therefore y=-2 $ From equation (i), $ x=5 $
$ \therefore $ Required point is $ (5,,-2) $ .
BETA
