QJEEM21Ma1f00: JEE Main 2021 Mathematics - Class 11 JEEM21Ma1f00 Solution: If P is a point on the parabola y = x 2 + 4 which is closest to …

Given, curve y = x² + 4

and, line y = 4x $-$ 1

Here, y = x² + 4


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

and y = 4x $-$ 1

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

Let the required point be P(x₁, y₁).

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

$\because$ Slopes will be equal.

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

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

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)