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\[{{x}^{2}}=2y\] \[\therefore \] \[{{h}^{2}}=2k\] ...(i) Let distance between\[(h,k)\]and (0,5) is D. \[\therefore \] \[D=\sqrt{{{h}^{2}}+{{(k-5)}^{2}}}\] \[=\sqrt{2k+{{(k-5)}^{2}}}\] On differentiating w.r.t.\[k,\]we get \[\frac{dD}{dk}=\frac{2+2(k-5)}{2\sqrt{2k+{{(k-5)}^{2}}}}\] For maxima and minima, put \[\frac{dD}{dk}=0\] \[\Rightarrow \] \[2+2(k-5)=0\] \[\Rightarrow \] \[k=4\] From Eq. (i), \[{{h}^{2}}=2\times 4=8\] \[\Rightarrow \] \[h=\pm 2\sqrt{2}\] Hence, the point at minimum distance is\[(\pm 2\sqrt{2},4)\].
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