Question

The point of parabola $ 2y={{x}^{2}}, $ which is nearest to the point $ (0,\,\,3) $ is

• A. $ (\pm 4,\,8) $
• B. $ (\pm 1,1/2) $
• C. $ (\pm 2,2) $
• D. None of these

Answer 1

Answer: (C)

Let $ (x,y) $ be any point on the parabola
$ 2y={{x}^{2}}. $
Let $ f(x)={{(x-0)}^{2}}+{{(y-3)}^{2}} $
$ ={{x}^{2}}+\left( \frac{{{x}^{2}}}{2}-3 \right) $
$ \Rightarrow $ $ f'(x)=2x+2{{\left( \frac{{{x}^{2}}}{2}-3 \right)}^{2}}
$ Put $ f'(x)=0 $
$ \Rightarrow $ $ 2x\left( 1+\frac{{{x}^{2}}}{2}-3 \right)=0 $ Now, $ f''(x)=2+2\left( \frac{{{x}^{2}}}{2}-3 \right)+2x(x) $
At $ x=0,\,f''(x)=2-6=-4 < 0, $
maximum At $ x=+2,\,f''(x) > 0, $
minimum At $ x=-2,\,\,f''(x) > 0, $ minimum
$ \therefore $ At $ x=\pm 2 $
$ \Rightarrow $ $ y=2 $
$ \therefore $ Required point is $ (\pm 2,2) $ .