JEE Main 2019 — Parabola Question with Solution

We know,

Equation of tangent to the parabola y² = 4ax is,

y = mx + $$\frac{a}{m}$$

$$\therefore$$  Equation of tangent to the parabola y² = 4x is,

y = mx + $$\frac{1}{m}$$

$$\Rightarrow$$  m²x $$-$$ ym + 1 = 0

This tangent is also the tangent to the circle x² + y² $$-$$ 6x = 0

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$$\therefore$$  $$\left|\frac{3m^2+1}{\sqrt{m^4+m^2}}\right|=3$$

$$\Rightarrow$$  (3m² + 1)² = 9(m⁴ + m²)

$$\Rightarrow$$  9m⁴ + 6m² + 1 = 9m⁴ + 9m²

$$\Rightarrow$$  3m² = 1

$$\Rightarrow$$  m = $$\pm$$ $$\frac{1}{\sqrt{3}}$$

So, possible tangents are

y = $$\frac{1}{\sqrt{3}}$$x + $$\sqrt{3}$$

$$\Rightarrow$$  $$\sqrt{3}$$y = x + 3

or   y = $$-$$ $$\frac{x}{\sqrt{3}}$$ $$-$$ $$\sqrt{3}$$

$$\Rightarrow$$  $$\sqrt{3}y$$ = $$-$$ x $$-$$ 3