The equation of tangent line to $y=2x^2+7$ which is parallel to the line $4x-y+3=0$ is :

$4x-y+1=0$ $4x-y+5=0$ $x+4y+5=0$ $4x-y+3=0$

$4x-y+5=0$

$y=2x^2+7$  ...(1)   line $4x-y+3=0$ differentiating wrt x   or $y=4x+3$ as $\frac{dy}{dx}=4$ so $\frac{dy}{dx}=4x=m=4$   slope of line (m) = 4 $⇒x=1$  from (1) so $y=2×1^2+7=9$ point of contact (1, 9) equation of tangent = $y-9=4(x-1)$ $y-9=4x-4$ so $4x-y+5=0$