# Tangents to parabola

1. Sep 21, 2008

### Mentallic

1. The problem statement, all variables and given/known data

I am given the parabola $$y=\frac{x^2}{2}$$

I need to find the equations of the 2 tangents to the parabola that pass through the point A(5,-2)

2. Relevant equations

$$y-y_{1}=m(x-x_{1})$$

$$a = \frac{1}{2}$$
therefore: tangents pass through the points $$P(p,\frac{p^{2}}{2})$$ and $$Q(q,\frac{q^{2}}{2})$$

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

3. The attempt at a solution

I began to check if the given point was outside the parabola
i.e. $$y_{1}<\frac{x^{2}_{1}}{2}$$

$$-2<\frac{5}{2}$$ therefore, the point lies outside the parabola and there are 2 lines that will pass through the point, and are a tangent to the parabola.

$$y+2=m(x-5)$$ where there are 2 values of m, each intersecting the parabola only once. i.e. tangent to parabola.

From here I am totally stumped. I can't use the 1st derivative as I don't know the x value for which the gradient will pass through the point.
Any help would be much appreciated.

Last edited: Sep 21, 2008
2. Sep 21, 2008

### Knissp

This means that your equation of your line must pass through two given points.

(5, -2) is one of them, so your equation $$y+2=m(x-5)$$ is a good place to start.

First, what is m? Remember, m is your slope...

Once you got that, now where does that second point have to be? If the line is tangent to the parabola with equation $$y=\frac{x^2}{2}$$...

3. Sep 21, 2008

### HallsofIvy

Staff Emeritus
You want to find (x, y) such that y= m(x- 5)- 2 and y'(x)= m. You should certainly be able to find the derivative of y in terms of x. Putting that in for y' gives three equations for x, y, and m. this will, of course, reduce to a quadratic equation so you can expect two solutions: two tangents.

4. Sep 21, 2008

### Mentallic

Thank you, I managed to solve it.

Last edited: Sep 22, 2008