Question

Find an equation of the line tangent to the curve y = 3x
2 − 1 and
parallel to the line 2x − y + 3 =

Answer 1

Given equation of curve is $y=3x^2-1$

$\therefore \frac{dy}{dx}=6x$

So, slope of tangent at any point $(x_1,y_1)$ is: $6x_1$

Now, according to question, this tangent is parallel to the given line $2x-y+3=0$

So, $6x_1=2\\$

$\Rightarrow x_1=\frac{1}{3}$

$\therefore y_1=3x_1^2-1=3(\frac{1}{3})^2-1=-\frac{2}{3}$

So, equation of tangent is: $y+\frac{2}{3}=2(x-\frac{1}{3})\\$

$\Rightarrow 3y+2=6x-2\\ \Rightarrow 6x-3y-4=0$