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

Question

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

Accepted Answer

Shortest distance between two  curve occurred along -

the common normal

Slope of normal to ${y^2} = x$ at point

$$P\left( {{t^2},t} \right)$$ is $-2t$ and

slope of line $y - x = 1$ is $1.$

As they are perpendicular to each other

$$\therefore$$ $$\left( { - 2t} \right) =  - 1 \Rightarrow t = {1 \over 2}$$

$$\therefore$$ $$P\left( {{1 \over 4},{1 \over 2}} \right)$$

and shortest distance $$ = \left| {{{{1 \over 2} - {1 \over 4} - 1} \over {\sqrt 2 }}} \right|$$

So shortest distance between them is $${{3\sqrt 2 } \over 8}$$

Solutions · 1