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 $A(0, 3)$

$\therefore AB= \sqrt{\left(\frac{K^{2}}{4}\right)^{2}+\left(K-3\right)^{2}}$
$\Rightarrow AB^{2}=\frac{K^{4}}{16}+K^{2}-6K+9$
$AB$ is shortest
$\therefore AB^{2}$ 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 $
$\Rightarrow \frac{d^{2}z}{dk^{2}}=\frac{3K^{2}}{4}+2$
$\Rightarrow \left(\frac{d^{2}Z}{dK^{2}}\right)_{k=2} >\,0$
$\because Z$ is minimum at $K=2$
$\therefore AB=\sqrt{\frac{16}{16}+\left(1\right)^{2}} =\sqrt{2}$