Question

The shortest distance between the line $x-y=1$ and the curve $x^2=2y$ is :

• A. $\frac{1}{2}$
• B. $0$
• C. $\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $\frac{1}{2\sqrt{2}}$

Shortest distance must be along common normal


$m_1\text{(slope of line}x-y=1)=1$
$\Rightarrow$ slope of perpendicular line $=-1$
Slope of tangent line at $(h,\frac{h^2}{2})$ is
$m_2=\frac{2x}{2}=x\Rightarrow m_2=h$
$\Rightarrow$slope of normal $=-\frac{1}{h}$
So, $-\frac{1}{h}=-1\Rightarrow h=1$
so point is $(1,\frac{1}{2})$
$\Rightarrow D=\left|\frac{1-\frac{1}{2}-1}{\sqrt{1+1}}\right|=\frac{1}{2\sqrt{2}}$