Applications Of Derivatives Question 367
Question: The point $ (0,,5) $ is closest to the curve $ x^{2}=2y $ at
[MNR 1983]
Options:
A) $ (2\sqrt{2},0) $
B) (0, 0)
C) $ (2,,2) $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
Let a point on the curve be (h, k)
Then $ h^{2}=2k $ …………..(i)
Distance = D = $ \sqrt{h^{2}+{{(k-5)}^{2}}} $ By (i); $ D=\sqrt{h^{2}+2k+{{(k-5)}^{2}}} $
$ \frac{dD}{dk}=\frac{1}{2\sqrt{2k+{{(k-5)}^{2}}}}\times 2(k-5)+2=0 $
$ x\in (-1,\infty ) $
So, at $ k=4 $ function D must be minimum.
Then point will be $ (\pm ,2\sqrt{2},,4) $ .
BETA
