Question

The point at which the normal to the curve y = `"x" + 1/"x", "x" > 0` is perpendicular to the line 3x – 4y – 7 = 0 is:

Options

• `(2, 5/2)`

• `(±2, 5/2)`

• `(-1/2, 5/2)`

• `(1/2, 5/2)`

Answer 1

`(2, 5/2)`

Explanation:

f(x) = `"x" + 1/"x", "x" > 0`

f'(x) = `1 - 1/"x"^2 = ("x"^2 - 1)/"x"^2, "x" > 0`

As normal to the curve y = f(x) at some point (x, y) is Ʇ to given line

`("x"^2/(1 - "x"^2)) xx 3/4` = −1 .....(m₁ × m₂ = −1)

x² = 4

x = ±2

But x > 0,

∴ x = 2

Therefore point = `(2, 5/2)`