Find equations for the two lines through the point (3, 13) that are tangent to the parabola $y = 6x - x^2$
since both lines must be tangent to the parabola their slope must be:
$$y' = 6- 2x$$
so.
$$6-2x = \frac{y-13}{x-3}$$ $$y = 12x-2x^2-5$$
but I did not get the lines what should I do?
You have to keep in mind that the numbers $x$ and $y$ are not arbitrary; they are such that $y=6x-x^2$. So, since $y$ is equal to both $6x-x^2$ and to $12x-2x^2-5$, you have $6x-x^2=12x-2x^2-5$. This is a quadratic equation, whose solutions are $1$ and $5$. It follows that the points of the parabola that you are after are $(1,5)$ and $(5,5)$.