Question

What normal to the curve $y=x^2$ forms the shortest chord?

• A. $y-\frac{1}{2}=\pm \frac{1}{\sqrt{2}}x+1.$
• B. $y+\frac{1}{2}=\pm \frac{1}{\sqrt{2}}x+1.$
• C. $y-\frac{1}{2}=\pm \frac{1}{\sqrt{2}}x-1.$
• D. $y-\frac{1}{2}=\pm \frac{1}{\sqrt{5}}x+1.$

Answer 1

Answer: (A)

$y-\frac{1}{2}=\pm \frac{1}{\sqrt{2}}x+1.$

Let $P(t,t^2)$ be the parametric co-ordinates of any point on the parabola $y=x^2.$

$\therefore \frac{dy}{dx}=2x=2t.$

Hence the normal at P is

$y-t^2=-\frac{1}{2t}(x-t)$ ...(1)

If it meets the curve again at Q whose parameter is say t' then the point $(t^{{}^{\prime }},t^{{}^{\prime }2})$ will satisfy (1).

$\therefore t^{{}^{\prime }2}-t^2=-\frac{1}{2t}(t^{{}^{\prime }}-t)$

or $\therefore t^{{}^{\prime }}+t=-\frac{1}{2t}(ast\ne t^{{}^{\prime }})$ ...(2)

$\therefore t^{{}^{\prime }}=-t-\frac{1}{2t}\cdot$

If l be the length of normal chord, then

$z=l^2=P{Q}^2=(t-t^{{}^{\prime }}{)}^2+(t^2-t^{{}^{\prime }2}{)}^2$

$z=(t-t^{{}^{\prime }}{)}^2[1+(t+t^{{}^{\prime }}{)}^2].$

Put for $t^{{}^{\prime }}$ from (2)

$\therefore z={(t+t+\frac{1}{2t})}^2[1+(\frac{-1}{2t}{)}^2]$

or $z=\frac{(4t^2+1{)}^2}{4t^2}\cdot \frac{4t^2+1}{4t^2}=\frac{(4t^2+1{)}^3}{16t^4}$

Now put $4t^2=p$ where $p$ is +ive.

$\therefore z=\frac{(p+1{)}^3}{p^2}=p+3+\frac{3}{p}+\frac{1}{p^2}$

$\frac{dz}{dp}=1-\frac{3}{p^2}-\frac{2}{p^3}=0$

or $p^3-3p-2=0or(p+1)(p^2-p-2)=0$

or $(p+1)(p+1)(p-2)=0$

$\therefore p=2asp\ne -1$

$\frac{d^{2}z}{d{p}^2}=\frac{6}{p^3}+\frac{6}{p^4}=+ive$ for $p=2=4t^2$

Hence z is minimum when $t^2=\frac{1}{2}$ or $t=\pm \frac{1}{\sqrt{2}}$

$z=l^2=\frac{(2+1{)}^3}{2^2}=\frac{27}{4}$

$\therefore l=\frac{3\sqrt{3}}{2}$

Hence from (1), the equations of normal are

$y-\frac{1}{2}=\pm \frac{1}{\sqrt{2}}x+1.$