Find the co-ordinates of points on the parabola $y = x^{2}$ closest to the point $A (0, a) .$

(\sqrt{\frac{2 a - 1}{2}}, \frac{2 a - 1}{2}) .

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(\sqrt{\frac{2 a + 1}{2}}, \frac{2 a + 1}{2}) .

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(- \sqrt{\frac{2 a - 1}{2}}, \frac{2 a - 1}{2}) .

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(- \sqrt{\frac{2 a + 1}{2}}, \frac{2 a + 1}{2}) .

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Solution

The correct options are
A $(\sqrt{\frac{2 a - 1}{2}}, \frac{2 a - 1}{2}) .$
C $(- \sqrt{\frac{2 a - 1}{2}}, \frac{2 a - 1}{2}) .$
Let $P (h, k)$ be any point on the parabola $x^{2} = y$
$∴ h^{2} = k$ ....(1)
Given point $A (0, a)$
Let $D = P A$
$z = D^{2} = h^{2} + {(k - a)}^{2}$
$\Rightarrow z = k + {(k - a)}^{2}$ (by (1))

Now, $\frac{d z}{d k} = 1 + 2 (k - a)$
For maxima or minima ,
$\frac{d z}{d k} = 0$
$∴ k = \frac{2 a - 1}{2}$
$\frac{d^{2} z}{d k^{2}} = 2$ i.e. positive
$∴ z$ has a minimum at $k = \frac{2 a - 1}{2}$
Also $h = \pm \sqrt{k}$
$∴$ P is $(h, k)$ i.e. $(\pm \sqrt{\frac{2 a - 1}{2}}, \frac{2 a - 1}{2}) .$

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Find the co-ordinates of points on the parabola y=x2 closest to the point A(0,a).

Find the co-ordinates of points on the parabola $y = x^{2}$ closest to the point $A (0, a) .$