Question

The points on the curve $ {{x}^{2}}=2y $ which are closest to the point $ (0,\,\,5) $ are

• A. $ (2,\,\,2),\,\,(-2,\,\,2) $
• B. $ (2\sqrt{2},\,\,4),\,\,(-2\sqrt{2},\,\,4) $
• C. $ (\sqrt{6},\,\,3),\,\,(-\sqrt{6},\,\,3) $
• D. $ (2\sqrt{3},\,\,6),\,\,(-2\sqrt{3},\,\,6) $

Answer 1

Answer: (B)

Let $ (h,\,\,k) $ be the point on the curve
$ {{x}^{2}}=2y $ $ ie, $ $ {{h}^{2}}=2k $ ... (i)
Let $ D $ be the distance between $ (h,\,\,k) $ and $ (0,\,\,5) $ .
$ \therefore $ $ D=\sqrt{{{h}^{2}}+{{(k-5)}^{2}}} $
$ =\sqrt{2k{{(k-5)}^{2}}} $
On differentiating w.r.t. $ k, $
we get $ \frac{dD}{dk}=\frac{2+2(k-5)}{\sqrt{2k+{{(k-5)}^{2}}}} $
For minima, put $ \frac{dD}{dk}=0 $
$ \Rightarrow $ $ 2+2(k-5)=0\Rightarrow k=4 $
$ \therefore $ From Eq. (i), we get $ {{h}^{2}}=2\times 4 $
$ \Rightarrow $ $ h=\pm 2\sqrt{2} $
Hence, closest point is $ (\pm 2\sqrt{2},\,\,4) $ .