Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

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