Question

On the parabola $y^2=64x,$ find the point nearest to the straight line $4x+3y-14=0$.

• A. $-24,9$
• B. $9,12$
• C. $-9,24$
• D. $9,-24$

Answer 1

Answer: (D)

$9,-24$

Given parabola is $y^2=64x$ ......... $\left(i\right)$

The line is $4x+3y-14=0$ ......... $\left(ii\right)$

Any point on the parabola will have the form $(\frac{t^2}{64},t)$

And points on the line will be of the form $(\frac{14-3t}{4},t)$

$\therefore$ Distance between $(\frac{t^2}{64},t)$ and $(\frac{14-3t}{4},t)$ is

$D=\sqrt{{(\frac{14-3t}{4}-\frac{t^2}{64})}^2+(t-t{)}^2}$

$D=\frac{14-3t}{4}-\frac{t^2}{64}$

For the nearest distance, $D^{\prime }=0$

$⟹-\frac{3}{4}-\frac{2t}{64}=0$

$⟹t=-24⟹y=-24$

At $y=-24,x=9$ on the parabola

Thus, the point nearest to the straight line is $(9,-24)$.