Question

The point on the curve \(x^{2}=2y\) which is nearest to the point\((0,5)\)is

• A. \((2\sqrt{2},4)\)
• B. \((2\sqrt{2},0)\)
• C. \((0,0)\)
• D. \((2,2)\)

Answer 1

The Correct Option is A

The given curve is \( x^{2}=2y.\)

For each value of x,the position of the point will be\((x,\frac{x^{2}}{2}).\)

The distance \(d(x)\)between the points\((x,\frac{x^{2}}{2}).\)and\((0,5)\)is given by,

\(d(x)=\sqrt{(x-0)^{2}+(\frac{x^{2}}{2}-5)^{2}}\)=\(\sqrt{x^{2}+\frac{x^{4}}{4}+25-5x^{2}}\)=\(\sqrt{\frac{x^{4}}{4}-4x^{2}+25}\)

\(∴d'(x)\)\(=\frac{(x^{3}-8x)}{2\sqrt{\frac{x^{4}}{4}-4x^{2}+25}}\)\(=\frac{(x^{3}-8x)}{\sqrt{x^{4}-16x^{2}+100}}\)

Now,\(d'(x)=0⇒x^{3}-8x=0\)

\(⇒x(x^{2}-8)=0\)

\(⇒x=0,\pm2\sqrt{2}\)

\(And,d''(x)=\)\(\frac{\sqrt{x^{4}-16x^{2}+100}(3x^{2}-8)-(x^{3}-8x).\frac{4x3-32x}{2\sqrt{x^{4}-16x^{2}+100}}}{(x^{4}-16x^{2}+100)}\)

\(=\frac{(x^{4}-16x^{2}+100)(3x^{2}-8)-2(x^{3}-8x)(x^{3}-8x)}{(x^{4}-16x^{2}+100)\frac{3}{2}}\)

\(=\frac{(x^{4}-16x^{2}+100)(3x^{2}-8)-2(x^{3}-8x^{2})2}{(x^{4}-16x^{2}+100)\frac{3}{2}}\)

When\(,x=0,then\space  d''(x)=\frac{36(-8)}{63}<0.\)

When\(,x=\pm2\sqrt{2},d''(x)>0.\)

By second derivative test,\(d(x)\) is the minimum at \(x=\pm2\sqrt{2}.\)

When\( x=\pm2\sqrt{2},y=\frac{(2\sqrt{2})2}{2}=4.\)

Hence,the point on the curve \(x^{2}=2y\)  which is nearest to the point\((0,5)\)is\((\pm2\sqrt{2},4).\)

The correct answer is A