JEE Main 2021 — Parabola Question with Solution

Given, curve y = x² + 4

and, line y = 4x $$-$$ 1

Here, y = x² + 4


$$\therefore$$ $$\frac{dy}{dx}=2x$$ ..... (i)

and y = 4x $$-$$ 1

$$\frac{dy}{dx}=4$$ ..... (ii)

Let the required point be P(x₁, y₁).

$$\therefore$$ $${\frac{dy}{dx}|}_P=2x_1$$ ..... (iii)

$$\because$$ Slopes will be equal.

$$\therefore$$ 2x₁ = 4 [from Eqs. (ii) and (iii)]

$$\Rightarrow x_1=\frac{4}{2}=2$$

Now, the given point P(x₁, y₁) lies on curve y = x² + 4,

$$\therefore$$ y₁ = x$${}_1^2$$ + 4

$$\Rightarrow$$ y₁ = 2² + 4 = 8

Hence, required coordinate of P = (2, 8)