Question

The shortest distance between the parabola $y^2=4x$ and the circle $x^2+y^2+6x-12y+20=0$ is

• A. $4\sqrt{2}-5$
• B. $0$
• C. $3\sqrt{2}+5$
• D. $1$

Answer 1

The correct option is A $4\sqrt{2}-5$

Equation of circle is $x^2+y^2+6x-12y+20=0$
$=(x+3{)}^2+(y-6{)}^2=25$
$\Rightarrow$ Centre $:(-3,6)$ and Radius $=5$
Differentiating $y^2=4x$ w.r.t $x$
$\Rightarrow \frac{dy}{dx}=\frac{2}{y}$
$\Rightarrow$Slope of normal to $y^2=4x$ at $(x_1,y_1)$ is $-\frac{y_1}{2}$
We know, the shortest distance is along the normal, so common normal to circle and parabola will pass through center of the circle $(-3,6)$
$\Rightarrow$ Equation of normal is
$(y-y_1)=\frac{-y_1}{2}(x-x_1)$
$\Rightarrow 6-y_1=\frac{-y_1}{2}(-3-x_1)$ $\{\because \text{passes through}(-3,6\left)\right\}$
$\Rightarrow 12=5y_1+x_1{y}_1$
$\Rightarrow 12=5y_1+\frac{y_1^3}{4}$
$\Rightarrow y_1^3+20y_1-48=0$
$\Rightarrow y_1=2\Rightarrow x_1=1$



$\Rightarrow$ Shortest distance will be distance between circle's centre and the point $(1,2)$ minus radius of the circle
$\Rightarrow \sqrt{(-3-1{)}^2+(6-2{)}^2}-$Radius
$\Rightarrow \text{S.D.}=4\sqrt{2}-5$