Question

If $Q$ is the point on the parabola $y^{2}=4 x$ that is nearest to the point $P(2,0)$, then $P Q=$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Let $Q$ be $(x, y)$
$\therefore P Q=\sqrt{(x-2)^{2}+(y-0)^{2}}$
$\Rightarrow P Q=\sqrt{(x-2)^{2}+y^{2}}$
$\Rightarrow P Q=\sqrt{(x-2)^{2}+4 x}\left[\because y^{2}=4 x\right]$
$\Rightarrow P Q=\sqrt{x^{2}+4}$
For minimum value of $P Q$
$\frac{d(P Q)}{d n}=0$
$\Rightarrow \frac{1}{2 \sqrt{x^{2}+4}}(2 x)=0$
$\Rightarrow x=0$
Also, $\frac{d^{2} P Q}{d x^{2}}=\frac{\sqrt{x^{2}+4}-x \cdot \frac{1}{2 \sqrt{x^{2}+4}} \cdot 2 x}{x^{2}+4}$
$\left.\therefore \frac{d^{2} P Q}{d x^{2}}\right|_{x=0}=1(+$ ive $)$
Hence, $P Q$ is minimum when $x=0$
$\therefore $ Minimum value of $P Q=\sqrt{0^{2}+4}=2$