Finding tangent lines to a parabola.

1. Oct 9, 2006

Jacobpm64

Using a graph to help you, find the equations of all lines through the origin tangent to the parabola
$$y=x^2-2x+4$$.

Sketch the lines on the graph.

I'm not sure how to start this off, I know the the first derivative would give me the slope of the tangent line at any point. If I find that:
$$y=x^2-2x+4$$
$$f'(x)=2x-2$$

Now, I know that my line has the slope $$2x-2$$ and it goes through the origin. So, I thought of plugging that into the slope-intercept form for a line, but it doesn't really work, I don't think:
$$0=(2x-2)(0)+b$$
$$b=0$$

The $$x$$'s go out, and I'm left with an equation of:
$$y=(2x-2)x$$
$$y=2x^2-2x$$

When I graphed this parabola and the original one, I noticed that the graphs intersected at the points that the tangent lines would go through. So, to find those points, I set the equations equal to each other:
$$x^2-2x+4=2x^2-2x$$
$$x^2-4=0$$
$$(x-2)(x+2)=0$$
$$x=2,x=-2$$

Since I just got the x-values, I plugged them in to the original function to get the corresponding y-values:
$$f(2)=(2)^2-2(2)+4=4$$
$$f(-2)=(-2)^2-2(-2)+4=12$$

Now, I can find the equations of the two tangent lines by using the points I found:
First tangent line:
(0, 0), (2, 4)
$$\frac{4-0}{2-0}=\frac{4}{2}=2$$
$$y=2x$$

Second tangent line:
(0, 0), (-2, 12)
$$\frac{12-0}{-2-0}=\frac{12}{-2}=-6$$
$$y=-6x$$

Is this correct? It seems like there would be an easier way to go about doing this... Does anyone know?... It seems like I made things work in some very weird, rough way.

Last edited: Oct 9, 2006