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Mathematics Question on Applications of Derivatives

The point on the curve x2=2yx^{2}=2y which is nearest to the point(0,5)(0,5)is

A

(22,4)(2\sqrt{2},4)

B

(22,0)(2\sqrt{2},0)

C

(0,0)(0,0)

D

(2,2)(2,2)

Answer

(22,4)(2\sqrt{2},4)

Explanation

Solution

The given curve is x2=2y. x^{2}=2y.

For each value of x,the position of the point will be(x,x22).(x,\frac{x^{2}}{2}).

The distance d(x)d(x)between the points(x,x22).(x,\frac{x^{2}}{2}).and(0,5)(0,5)is given by,

d(x)=(x−0)2+(x22−5)2d(x)=\sqrt{(x-0)^{2}+(\frac{x^{2}}{2}-5)^{2}}=x2+x44+25−5x2\sqrt{x^{2}+\frac{x^{4}}{4}+25-5x^{2}}=x44−4x2+25\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⇒x3−8x=0d'(x)=0⇒x^{3}-8x=0

⇒x(x2−8)=0⇒x(x^{2}-8)=0

⇒x=0,±22⇒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)}

=(x4−16x2+100)(3x2−8)−2(x3−8x)(x3−8x)(x4−16x2+100)32=\frac{(x^{4}-16x^{2}+100)(3x^{2}-8)-2(x^{3}-8x)(x^{3}-8x)}{(x^{4}-16x^{2}+100)\frac{3}{2}}

=(x4−16x2+100)(3x2−8)−2(x3−8x2)2(x4−16x2+100)32=\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  d′′(x)=36(−8)63<0.,x=0,then\space d''(x)=\frac{36(-8)}{63}<0.

When,x=±22,d′′(x)>0.,x=\pm2\sqrt{2},d''(x)>0.

By second derivative test,d(x)d(x) is the minimum at x=±22.x=\pm2\sqrt{2}.

Whenx=±22,y=(22)22=4. x=\pm2\sqrt{2},y=\frac{(2\sqrt{2})2}{2}=4.

Hence,the point on the curve x2=2yx^{2}=2y which is nearest to the point(0,5)(0,5)is(±22,4).(\pm2\sqrt{2},4).

The correct answer is A