Find the point on the curve which is nearest to the point (2,-8).

Find the point on the curve $y^{2}= 4x,$ which is nearest to the point (2,-8).

Answers (1)

given the curve of the form $y^{2}= 4x$ ,which is a means to the point $\left ( 2,-8 \right )$ and let $p\left ( x,y \right )$ be the point on the curve.
$\because y^{2}= 4x$
then $\because \frac{y^{2}}{4}= x$
$\therefore p\left ( x,y \right ) will \: be\: p\left ( \frac{y^{2}}{4},y \right )$
Now the distance between point A & P is given by
$AP= \sqrt{\left ( x-2 \right )^{2}+\left ( y+8 \right )^{2}}$
$AP= \sqrt{\left ( \frac{y^{2}}{4} -2\right )^{2}+\left ( y+8 \right )^{2}}$
$\Rightarrow \sqrt{\frac{y^{2}}{16}-y^{2}+4+y^{2}+16y+64}\Rightarrow \sqrt{\frac{y^{4}}{16}+16y+68}$ and Let Z $z= \frac{y^{4}}{16}+16y+68-(i)$
$\frac{dz}{dy}= \frac{1}{16}\times 4y^{3}+16$
$= \frac{y^{3}}{4}+16$
For the maximum or minimum value of Z, we have
$\frac{dz}{dy}= 0$
$\Rightarrow \frac{y^{3}}{4}+16= 0 \: \Rightarrow y^{3}+64= 0$
$\Rightarrow \left ( y+4 \right )\left ( y^{2} -4y+16\right )= 0 \left ( \because y^{2}-4y+16= 0\: gives\:imaginary\: value \: of\: y \right )$ $\Rightarrow y= -4$
Now, $\frac{d^{2}z}{dy^{2}}= \frac{1}{4}\times 3y^{2}= \frac{3}{4}y^{2}$
At $y= -4$
$\frac{d^{2}Z}{dy^{2}}= \frac{3}{4}\left ( -4 \right )^{2}= 12> 0$
Thus, z is maximum when $y= -4$
substituting $y= -4$ in the equation of the curve $y^{2}= 4x$
we have $x= 4$
Hence, the point(4,-4) on the curve $y^{2}= 4x$ is nearest to the point(12,-8)