Question

The point $(0,5)$ is closest to the curve $x^2=2y$ at

• A. $(2\sqrt{2},0)$
• B. $(0,0)$
• C. $(2,2)$
• D. $noneofthese$

Answer 1

Answer: (D)

$noneofthese$

Let the parametric point on the parabola be
$P=(2t,2t^2)$ and let $K=(0,5)$
Hence
$P{K}^2=m$
$=4t^2+(2t-5{)}^2$
$=4t^2+4t^4-20t^2+25$
$=4t^4-16t^2+25$
Hence
$\frac{dm}{dt}$
$=16t^3-32t$
$=16t(t^2-2)$
$=0$
Hence $t=0$ or $t=\pm \sqrt{2}$
Now
$\frac{d^{2}m}{d{t}^2}$
$=48t^2-32$
Hence
${\frac{d^{2}m}{d{t}^2}}_{t=\pm \sqrt{2}}>0$ ... minima
Hence
$t=-\sqrt{2}$ and $t=\sqrt{2}$
Hence the points are $(2\sqrt{2},4)$ and $(-2\sqrt{2},4)$