How to find equation of tangent line to $x^2 = 2y$ at $(-3, 9/2)$ The equation of the parabola is $x^2 = 2y$. The point where where the tangent line touches the parabola is $(-3, 9/2)$. What is the equation of the tangent line and its $x$-intercept?
This is my review work for a test tomorrow, so far this is the only part I do not understand. 
You are supposed to set up an isosceles triangle and in that way solve the problem, but honestly I do not know how to even set it up
Please help
 A: Here is a hint. This tangent line must pass through $(-3,9/2)$, so it has the equation
$$y = m(x+3) - 9/2,$$
where $m$ is the slope of the tangent line.
But the tangent line also must intersect the parabola in exactly one point, so the system of the above equation and $y = x^2/2$ must have exactly one solution. Plugging in for $y$ from the second equation, you end up with
$$x^2/2 = m(x+3) - 9/2$$
or equivalently
$$x^2 - 2mx -6m + 9 = 0.$$
Recall that the quadratic equation $ax^2+bx+c=0$ has $(0,1,2)$ (real) solutions if the discriminant $\mathcal{D} = b^2-4ac$ is $(<,=,>) 0$. You get one solution if and only if $b^2-4ac=0$ or equivalently $b^2 = 4ac$. In our case, $a=1,b=-2m,c=-6m+9$. So, we have exactly one solution if the discriminant $b^2-4ac$ is zero, in other words,
$$(-2m)^2 = 4 \cdot 1 \cdot (-6m+9).$$
You should be able to take it from here - just solve the resulting quadratic equation.
EDIT inserted explanation about the discriminant.
A: Let $P=(p,p^2/2)$ be a point on the parabola. Let $q=(q,q^2/2)$ be an other point on the parabola. Then the line $PQ$ is 
$$
\frac{y-q^2/2}{x-q}=\frac{y-p^2/2}{x-p},
$$
or,
$$
y-\frac{p+q}{2}x+\frac{pq}{2}=0.
$$
Now, let $q$ close to $p$ we get the tangent
$$
y-px+\frac{p^2}{2}=0.
$$
Since $(-3,9/2)$ is a point on the parabola, $p=-3$...
