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
$\frac{d^{2}D}{dk^{2}} = 2 + 12 k^2$
$\left[\frac{d^{2}D}{dk^{2}}\right]_{k=1} =14>0$
Hence, at $k = 1$, the distance is minimum.
Shortest distance $ = \sqrt{ (1-3)^2 + (1^2)^2} = \sqrt{5}$