JEE Main 2019 — Parabola Question with Solution

Slope of the line y = x is 1.

$$\therefore$$ The shortest distance between line y = x and parabola y² = x – 2 is the distance between line y = x and tangent of parabola having slope 1.

Slope of tangent at point P on the parabola y² = x – 2 is,

2y$$\frac{dy}{dx}$$ = 1

$$\Rightarrow$$ $$\frac{dy}{dx}$$ = $$\frac{1}{2y}$$ = slope

$$\therefore$$ $$\frac{1}{2y}$$ = 1

$$\Rightarrow$$ y = $$\frac{1}{2}$$

$$\therefore$$ x coordinate of point P is,

$$\frac{1}{4}$$ = x – 2

$$\Rightarrow$$ x = $$\frac{9}{4}$$

Shortest distance = length of perpendicular from P on line x – y = 0

= $$\frac{\left|\frac{9}{4}-\frac{1}{2}\right|}{\sqrt{1^2+1^2}}$$ = $$\frac{7}{4\sqrt{2}}$$