# 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

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Do you know calculus? – John Marty May 2 '13 at 20:50
No sir I do not, I am in PreCalculus. I have looked up the problem and I keep finding things about using derivatives , but we have not learned those yet. I will learn about them next year. – Steven May 2 '13 at 21:01
To be honest, the only way I know to find the gradient is to use derivatives, once you have found the gradient the problem can be solved with simultaneous equations. – John Marty May 2 '13 at 21:14
What you can do, however, is draw an accurate diagram, draw an accurate tangent (at the given point), and then calculate the gradient. With a stress on being VERY precise. Not sure about bearing of drawing an isosceles triangle on this problem. – John Marty May 2 '13 at 21:15
Were you taught a special relationship between the vertices of an isosceles triangle with $2$ sides tangent to a parabola? Hint: With the information you have, the $y$-intercept of this triangle can be found by inspection. – John Douma May 2 '13 at 22:14

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.

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Ahha, I follow and understand up until you say the discriminant is zero. Why does that matter, or how does that relate to the problem? Also, how do you go from the x2−2mx−6m+9=0 step to the (−2m)2=4⋅1⋅(−6m+9) step? – Steven May 2 '13 at 22:27
@Steven Please see the edit which explains the concept of the discriminant. – gt6989b May 3 '13 at 4:03

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, $p=-3$...

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+1, this is clever :) – gt6989b May 3 '13 at 4:05