Question

The shortest distance from $(0,3)$ to the parabola $y^2=4x$ is

• A. $2$
• B. $\sqrt{2}$
• C. $5$
• D. $\sqrt{5}$

Answer 1

Answer: (B)

Let a point $B$ on parabola is
$B\left(\frac{K^2}{4},K\right)$ and $A(0,3)$

$\therefore AB=\sqrt{{\left(\frac{K^2}{4}\right)}^2+{\left(K-3\right)}^2}$
$\Rightarrow A{B}^2=\frac{K^4}{16}+K^2-6K+9$
$AB$ is shortest
$\therefore A{B}^2$ is also shortest
Let $AB=Z$
$\therefore Z=\frac{K^4}{16}+K^2-6K+9$
$\Rightarrow \frac{dZ}{dK}=\frac{K^3}{4}+2K-6$
For maxima or minima
put $\frac{dZ}{dK}=0$
$\Rightarrow \frac{K^2}{4}+2k-6=0$
$\Rightarrow K^2+8K-24=0$
$\Rightarrow \left(K-2\right)\left(K^2+2K+12\right)=0$
$K=2$
$\Rightarrow \frac{d^{2}z}{d{k}^2}=\frac{3K^2}{4}+2$
$\Rightarrow {\left(\frac{d^{2}Z}{d{K}^2}\right)}_{k=2}>0$
$\because Z$ is minimum at $K=2$
$\therefore AB=\sqrt{\frac{16}{16}+{\left(1\right)}^2}=\sqrt{2}$