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

The correct option is A $(9,-24)$

Given equation of parabola is
$y^2=64x$ ......(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 equation (i), we get
$2y\frac{dy}{dx}=64$
$\Rightarrow \frac{dy}{dx}=\frac{32}{y}$
Also, slope of the given line is $-\frac{4}{3}$
$\therefore -\frac{4}{3}=\frac{32}{y}\Rightarrow y=-24$
From equation (i), ${(-24)}^2=64x\Rightarrow x=9$
$\therefore$ the required point is $(9,-24)$