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Q.
The minimum distance from the point $(4, 2)$ to the parabola $y^2=8x $ is
Application of Derivatives
Solution:
Let $d =$ distance of a point $(x, y)$, of the parabola from the point $(4, 2)$
$\therefore D=d^{2}=\left(x-4\right)^{2}+\left(y-2\right)^{2}$
$=x^{2}+16-8x+y^{2}-4y+4$
$\left[\because y^{2}=8x\right]$
$=\frac{y^{4}}{64}+16-y^{2}-4y+4$
$=\frac{y^{4}}{64}-4y+20$
$\therefore \frac{d}{dy}\left(D\right)=\frac{4\,y^{3}}{64}-4$ and $\frac{d^{2}}{dy^{2}}\left(D\right)=\frac{12y^{2}}{64} > 0$
$\therefore D$ is Min when $\frac{d\left(D\right)}{dy}$
$\Rightarrow \frac{4y^{3}}{64}-4=0$
$\Rightarrow y^{3}=64$
$\Rightarrow y=4$.
But $y^{2}=8x$
$\therefore 16=8x$
$\Rightarrow x=2$.
$\therefore $ reqd. min. distance
$=\sqrt{\left(2-4\right)^{2}+\left(4-2\right)^{2}}$
$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$