The equation of tangent line to $y=2x^2+7$ which is parallel to the line $4x-y+3=0$ is :
Answer & explanation
Correct answer: option 2
| $y=2x^2+7$ ...(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 $4x-y+5=0$