Determine the points on the curve x^2 = 4y which are nearest to t… - Vidyadip

Question

Determine the points on the curve $x^2 = 4y$ which are nearest to the point $(0, 5).$

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Answer

Let the point $(x, y)$ on the curve $x^2 = 4y$ be nearest to $(0, 5),$ Then
$x^2 = 4y$
$\Rightarrow\text{y}^{2}=\frac{\text{x}^{2}}{\text{a}}\ ...(\text{i})$
Also, $\text{d}^{2}=(\text{x})^2+(\text{y}-5)^{2}$ [using distance formula]
Now, $\text{Z}=\text{d}^{2}=(\text{x})^{2}+(\text{y}-5)^{2}$
$\text{Z}=(\text{x})^{2}+\Big(\frac{\text{x}^{2}}{\text{a}}-5\Big)$
$\text{Z}=\text{x}^{2}+\frac{\text{x}^{4}}{16}+25-\frac{5\text{x}^{2}}{2}$
$\Rightarrow \frac{\text{dZ}}{\text{dy}}=2\text{x}+\frac{4\text{x}^{3}}{16}-5\text{x}$
For maximum or minimum value of Z, we must have $\frac{\text{dZ}}{\text{dy}}=0$
$\Rightarrow 2\text{x}+\frac{4\text{x}^{3}}{16}-5\text{x}=0$
$\Rightarrow \frac{4\text{x}^{3}}{16}=3\text{x}$
$\Rightarrow \text{x}=\pm2\sqrt{3}$
Substiting the value of x in eq.(i), we get $y = 3$
Now, $\frac{\text{d}^{2}\text{Z}}{\text{dy}^{2}}=2+\frac{12\text{x}^{2}}{16}-5$
$\frac{\text{d}^{2}\text{Z}}{\text{dy}^{2}}=9-6$
$= 6 > 0$
So, the required nearest point is $(\pm2\sqrt{3}, 3)$

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