Question

The point on the parabola $y^2 = 64x$ which is nearest to the line $4x + 3y + 35 = 0$ has coordinates

• A. (9, -24)
• B. (1, 81)
• C. (4, -16)
• D. (-9, -24)

Answer 1

Answer: (A)

(9, -24)

Answer 2

Given equation of parabola is
$y^{2}=64 x\,\,\,...(i)$
The point at which the tangent to the curve is parallel to the line is the nearest point on the curve.
On differentiating both sides of E (i), we get
$2 y \frac{d y}{d x} =64$
$\Rightarrow \frac{d y}{d x} =\frac{32}{y}$
Also, slope of the given line is $-\frac{4}{3}$.
$\therefore -\frac{4}{3}=\frac{32}{y}$
$\Rightarrow y=-24$
From E (i), $(-24)^{2}=64 x$
$\Rightarrow x=9$
$\therefore$ Hence, the required point is $(9,-24) .$