The point on the curve $y=8x-x^2$, where the tangent is parallel to the line $8x-4y -1=0$ is :

(3, 15)

The correct answer is Option (1) → (3, 15)

Rearrange the given equation into the slope-intercept form,

$8x-4y-1=0$   $[y=mx+c]$

$⇒y=2x-\frac{1}{4}$

$⇒m=2$

$y=8x-x^2$ [Given]

$\frac{dy}{dx}=8-2x$

$⇒8-2x=2$

$⇒2x=6$

$⇒x=3$

$⇒y=8(3)-(3)^2=15$

$⇒(3,15)$