Parabola tangents that form angle with given line

Question

Given parabola $y^2=12x$ and line $y=3x-4$, find all tangents of parabola that form $45^\circ ( $or $ 135 ^\circ )$ angle with given line.

My attempt: consider $f(y)=\frac{y^2}{12} $, then $f'(y)=\frac{y}{6}$. Now since $tan(45^\circ)=1$, it follows that $1=|\frac{\frac{y}{6}-3}{1+3\frac{y}{6}}|$, and this gets me $y=-12$ and $y=3$, for which tangent lines do not form $45^\circ$ angle with that line.

What am I doing wrong?

Answer 1

Let $m$ is a slop of the tangent.

Thus, since $\tan45^{\circ}=1$, we obtain: $$\left|\frac{m-3}{1+3m}\right|=1,$$ which gives $m-3=1+3m$ and $m=-2$ or

$m-3=-1-3m$ and $m=\frac{1}{2}$.

$y^2=2px$ is an equation of the parabola. Thus, $p=6$.

In another hand, $yy_1=p(x+x_1)$ is an equation of the tangent in the point $(x_1,y_1)$.

Thus, $yy_1=6(x+x_1)$ is an equation of our tangent.

1. $m=-2$.

Thus, $\frac{6}{y_1}=-2$, which gives $y_1=-3$ and $x_1=\frac{3}{4}$ and we get an equation og the tangent: $$y+3=-2\left(x-\frac{3}{4}\right)$$ or $$y=-2x-\frac{3}{2}.$$ 2. $m=\frac{1}{2}$.

Thus, $\frac{6}{y_1}=\frac{1}{2}$, which gives $y_1=12$ and from here $x_1=12$.

Hence, an equation of the tangent is $$y-12=\frac{1}{2}(x-12)$$ or $$y=\frac{1}{2}x+6.$$ Done!