Question

The nearest point on the curve $y^2=4x$ to $(2,1)$ is

• A. (4,4)
• B. (1,2)
• C. (9,6)
• D. (4,5)

Answer 1

Answer: (B)

Let $P(x,y)$ be any point on the curve $y^2=4x$.
$S=\sqrt{(x-2{)}^2+(y-1{)}^2}$
$\Rightarrow S^2=(x-2{)}^2+(y-1{)}^2={\left[\frac{y^2}{4}-2\right]}^2+(y-1{)}^2=f(y)$ (say)
$\Rightarrow f^{\prime }(y)=2\left[\frac{y^2}{4}-2\right]\left[\frac{2y}{4}\right]+2(y-1)$
$=\frac{\left(y^2-8\right)y+8(y-1)}{4}=\frac{y^3-8y+8y-8}{4}=\frac{y^3-8}{4}$
$\Rightarrow f^{\prime \prime }(y)=\frac{3y^2}{4}$
$f^{\prime }(y)=0$
$\Rightarrow \frac{y^3-8}{4}=0$
$\Rightarrow y^3-8=0$
$\Rightarrow y=2$
$f^{\prime \prime }(2)=\frac{3.2^2}{4}=3>0$.
When $y=2$, the distance between the points is minimum.
$\therefore y=2$
$\Rightarrow x=1.$
$\therefore$ Required point $=(1,2)$.