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If $Q$ is the point on the parabola $y^2=4 x$ that is nearest to the point $P(2,0)$, then $P Q=$
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Verified Answer
The correct answer is:
2
Let $Q$ be $(x, y)$
$$
\begin{array}{lll}
\therefore & P Q=\sqrt{(x-2)^2+(y-0)^2} & \\
\Rightarrow & P Q=\sqrt{(x-2)^2+y^2} & \\
\Rightarrow & P Q=\sqrt{\left(x-4^2+4 x\right.} & {\left[\because y^2=4 x\right]} \\
\Rightarrow & P Q=\sqrt{x^2+4} &
\end{array}
$$
For minimum value of $P Q$
$$
\frac{d(P Q)}{d n}=0 \Rightarrow \frac{1}{2 \sqrt{x^2+4}}(2 x)=0 \Rightarrow x=0
$$
AIso, $\frac{d^2 P Q}{d x^2}=\frac{\sqrt{x^2+4}-x \cdot \frac{1}{2 \sqrt{x^2+4}} \cdot 2 x}{x^2+4}$
$$
\left.\therefore \quad \frac{d^2 P Q}{d x^2}\right|_{x_0}=1(+ \text { ive })
$$
Hence, $P Q$ is minimum when $x=0$
$\therefore$ Minimum value of $P Q=\sqrt{0^2+4}=2$
$$
\begin{array}{lll}
\therefore & P Q=\sqrt{(x-2)^2+(y-0)^2} & \\
\Rightarrow & P Q=\sqrt{(x-2)^2+y^2} & \\
\Rightarrow & P Q=\sqrt{\left(x-4^2+4 x\right.} & {\left[\because y^2=4 x\right]} \\
\Rightarrow & P Q=\sqrt{x^2+4} &
\end{array}
$$
For minimum value of $P Q$
$$
\frac{d(P Q)}{d n}=0 \Rightarrow \frac{1}{2 \sqrt{x^2+4}}(2 x)=0 \Rightarrow x=0
$$
AIso, $\frac{d^2 P Q}{d x^2}=\frac{\sqrt{x^2+4}-x \cdot \frac{1}{2 \sqrt{x^2+4}} \cdot 2 x}{x^2+4}$
$$
\left.\therefore \quad \frac{d^2 P Q}{d x^2}\right|_{x_0}=1(+ \text { ive })
$$
Hence, $P Q$ is minimum when $x=0$
$\therefore$ Minimum value of $P Q=\sqrt{0^2+4}=2$
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