A) \[(2,\,\,2),\,\,(-2,\,\,2)\]
B) \[(2\sqrt{2},\,\,4),\,\,(-2\sqrt{2},\,\,4)\]
C) \[(\sqrt{6},\,\,3),\,\,(-\sqrt{6},\,\,3)\]
D) \[(2\sqrt{3},\,\,6),\,\,(-2\sqrt{3},\,\,6)\]
Correct Answer: B
Solution :
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\[(\pm 2\sqrt{2},\,\,4)\].
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