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The point ${ }^{(0,5)}$ is closest to the curve $x^2=2 y$ at
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Let a point on the curve by $(h, k)$
Then $h^2=2 k...(i)$
$\begin{aligned}& \text { Distance }=D=\sqrt{h^2+(k-5)^2} \\
& \text { By (i); } D=\sqrt{2 k+(k-5)^2} \\
& \frac{d D}{d k}=\frac{1}{2 \sqrt{2 k+(k-5)^2}} \times 2(k-5)+2=0 \quad \Rightarrow k=4\end{aligned}$
So, at $k=4$ function $D$ must be minimum.
Then point will be $( \pm 2 \sqrt{2}, 4)$.
Then $h^2=2 k...(i)$
$\begin{aligned}& \text { Distance }=D=\sqrt{h^2+(k-5)^2} \\
& \text { By (i); } D=\sqrt{2 k+(k-5)^2} \\
& \frac{d D}{d k}=\frac{1}{2 \sqrt{2 k+(k-5)^2}} \times 2(k-5)+2=0 \quad \Rightarrow k=4\end{aligned}$
So, at $k=4$ function $D$ must be minimum.
Then point will be $( \pm 2 \sqrt{2}, 4)$.
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