Find the point on the parabolas

Question:

Find the point on the parabolas $x^{2}=2 y$ which is closest to the point $(0,5)$.

Solution:

Let the required point be $(x, y)$. Then,

$x^{2}=2 y$

$\Rightarrow y=\frac{x^{2}}{2}$          ....(1)

The distance between points $(x, y)$ and $(0,5)$ is given by

$d^{2}=(x)^{2}+(y-5)^{2}$

Now,

$d^{2}=Z$

$\Rightarrow Z=(x)^{2}+\left(\frac{x^{2}}{2}-5\right)^{2}$

$\Rightarrow Z=x^{2}+\frac{x^{4}}{4}+25-5 x^{2}$

$\Rightarrow \frac{d Z}{d y}=2 x+x^{3}-10 x$

For maximum or a minimum values of $Z$, we must have

$\frac{d Z}{d y}=0$

$\Rightarrow x^{3}-8 x=0$

$\Rightarrow x^{2}=8$

$\Rightarrow x=\pm 2 \sqrt{2}$

Substituting the value of $x$ in eq. (1), we get

$y=4$

$\frac{d^{2} Z}{d y^{2}}=3 x^{2}-8$

$\Rightarrow \frac{d^{2} Z}{d y^{2}}=24-8=16>0$

So, the nearest point is $(\pm 2 \sqrt{2}, 4)$.

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