Question

If $P$ is a point on the parabola $y=x^2+4$ which is closest to the straight line $y=4x-1$, then the co-ordinates of $P$ are

• A. (3,13)
• B. (1,5)
• C. (-2,8)
• D. (2,8)

Answer 1

Answer: (D)

$P:y=x^2+4$
$k=h^2+4$
$L:y=4x-1$
$y-4x+1=0$
$d=AB=\left|\frac{k-4h+1}{\sqrt{5}}\right|$
$=\left|\frac{h^2-4-4h+1}{\sqrt{5}}\right|$
$\frac{d(d)}{dh}=\frac{2h-4}{\sqrt{5}}=0$
$h=2$
$\frac{d^2(d)}{d{h}^2}=\frac{2}{\sqrt{5}}>0$
$\therefore k=4+4=8$
$\therefore$ Point (2,8)