Find the co-ordinates of the point on the curve $4 y = x^{2}$ which are nearest to the point $(0, 5)$ .

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Solution

Let point $(x_{1}, y_{1})$
$4 y_{1} = x_{1}^{2}$ ..........(i)
$d = \sqrt{(x_{1} - 0)^{2} + (y_{1} - 5)^{2}}$
$d^{2} = x_{1}^{2} + y_{1}^{2} + 25 - 10 y_{1}$
$d^{2} = 4 y_{1} + y_{1}^{2} + 25 - 10 y_{1}$
Differentiate
$2 d d^{'} = 4 + 2 y_{1} - 10$
For maximum & minimum
$4 + 2 y_{1} - 10 = 0$
$2 y_{1} = 6$
$y_{1} = 3$
$4 \times 3 = x_{1}^{2}$
$x_{1} = \sqrt{12}$

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