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The point on the curve $x^2=2 y$ which is nearest to the point $(0,5)$ is
(a) $(2 \sqrt{2}, 4)$
(b) $(2 \sqrt{2}, 0)$
(c) $(\mathbf{0}, 0)$
(d) $(2,2)$
The point on the curve $x^2=2 y$ which is nearest to the point $(0,5)$ is
(a) $(2 \sqrt{2}, 4)$
(b) $(2 \sqrt{2}, 0)$
(c) $(\mathbf{0}, 0)$
(d) $(2,2)$
Solution:
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Verified Answer
(a) Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be a point on the curve The other point is $\mathrm{A}(0,5)$
$\mathrm{Z}=\mathrm{PA}^2=\mathrm{x}^2+\mathrm{y}^2+25-10 \mathrm{y}\left[\because \mathrm{x}^2=2 \mathrm{y}\right]$
$z=y^2-8 y+25 \therefore \frac{d z}{d y}=2 y-8, \frac{d^2 z}{d y^2}=2=+v e$
$\begin{aligned}
&\frac{\mathrm{dz}}{\mathrm{dy}}=0 \Rightarrow \mathrm{y}=4 \frac{\mathrm{d}^2 \mathrm{z}}{\mathrm{dy}^2}=+\mathrm{ve}, \mathrm{z} \text { is minimum } \\
&\therefore \mathrm{x}^2=2 \mathrm{y}=2 \times 4=8 \\
&\therefore \mathrm{x}=2 \sqrt{2} \Rightarrow \mathrm{z} \text { is minimum at }(2 \sqrt{2}, 4) \\
&\Rightarrow \sqrt{\mathrm{z}} \text { is minimum at }(2 \sqrt{2}, 4)
\end{aligned}$
$\mathrm{Z}=\mathrm{PA}^2=\mathrm{x}^2+\mathrm{y}^2+25-10 \mathrm{y}\left[\because \mathrm{x}^2=2 \mathrm{y}\right]$
$z=y^2-8 y+25 \therefore \frac{d z}{d y}=2 y-8, \frac{d^2 z}{d y^2}=2=+v e$
$\begin{aligned}
&\frac{\mathrm{dz}}{\mathrm{dy}}=0 \Rightarrow \mathrm{y}=4 \frac{\mathrm{d}^2 \mathrm{z}}{\mathrm{dy}^2}=+\mathrm{ve}, \mathrm{z} \text { is minimum } \\
&\therefore \mathrm{x}^2=2 \mathrm{y}=2 \times 4=8 \\
&\therefore \mathrm{x}=2 \sqrt{2} \Rightarrow \mathrm{z} \text { is minimum at }(2 \sqrt{2}, 4) \\
&\Rightarrow \sqrt{\mathrm{z}} \text { is minimum at }(2 \sqrt{2}, 4)
\end{aligned}$
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