Question

The shortest distance from the point $(3,0)$ to the parabola $y=x^2$ is.

• A. $\sqrt{3}$
• B. 3
• C. 5
• D. $\sqrt{5}$

Answer 1

Answer: (D)

Let $(k,k^2)$ be any point on the parabola. $d^2=(k-3{)}^2+(k^2-0{)}^2$
$D=(k-3{)}^2+k^4($ where $d^2=D)$
differentiating w.r.t. k
$\frac{dD}{dk}=2(k-3)+4k^3$....(i)
For maximum or minimum $\frac{dD}{dk}=0$
$\Rightarrow 2k-6+4k^3=0\Rightarrow 2k^3+k-3=0$
$\Rightarrow (k-1)(2k^3+2k+3)=0$
$\Rightarrow (k-1)=0$ or $2k^2+2k+3=0\Rightarrow k=1$
Again differentiating·(i) w.r.t. k
$\frac{d^{2}D}{d{k}^2}=2+12k^2$
${\left[\frac{d^{2}D}{d{k}^2}\right]}_{k=1}=14>0$
Hence, at $k=1$, the distance is minimum.
Shortest distance $=\sqrt{(1-3{)}^2+(1^2{)}^2}=\sqrt{5}$