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