Find Tangents to Parabola Passing Through Point A (5,-2)

[Mentallic]

Homework Statement

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)

Homework 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

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.

 

[Knissp]

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

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}...

 

[HallsofIvy]

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.

 

[Mentallic]

Thank you, I managed to solve it.