# Tangent to a parabola

• Dec 16th 2007, 03:18 AM
free_to_fly
Tangent to a parabola
Got stuck on another hw problem, can someone help please?

Show that the line with the equation 4y-3x=20 is a tangent to both the circle with equation x^2+y^2=16 and the parabola with equation y^2=15x.

I don't know which points the tangent touch the circle and the parabola at, and I'm not sure how to find it.
• Dec 16th 2007, 04:43 AM
mathdeity
Someone pointed in out in another thread of ours that it is only necessary to show that there is a single point of intersection to prove tangency. In this case, I think the easiest approach is to take 4y-3x=20 and isolate one of the variables. Next, substitute your result into the circle and find the single point of tangency.

Repeat the same process with 4y-3x=20 and the parabola.
• Dec 16th 2007, 06:32 AM
galactus
The line equation is $\displaystyle y=\frac{3}{4}x+5$

We have the circle $\displaystyle y=\pm\sqrt{16-x^{2}}$ and the parabola

$\displaystyle y=\pm\sqrt{15x}$

Set the line equation equal to the circle and the parabola, respectively:

$\displaystyle \frac{3}{4}x+5=\sqrt{16-x^{2}}$

Solving for x, we see $\displaystyle x=\frac{-12}{5}$

$\displaystyle \frac{3}{4}x+5=\sqrt{15x}$

$\displaystyle x=\frac{20}{3}$

Now, find the derivative(slope) of the circle and parabola at those points.

If they both equal 3/4(the slope of the line), then we're in business.
• Dec 16th 2007, 06:40 AM
free_to_fly
I've done that question now, thanks for the help.

But I've come across another similar question that doesn't give the equation of the tangent or the point of intersection, so I'm not sure what to do again:

Find an equation of the line which is tangent to both the parabola y^2=4ax and the parabola x^2=4ay.

The answer is x+y+a=0, but I've got no idea where that came from. Can I assume that the tangent touches the two parabola's at the same point?

Help would be greatly appreciated.
• Dec 16th 2007, 11:32 AM
topsquark
Quote:

Originally Posted by mathdeity
Someone pointed in out in another thread of ours that it is only necessary to show that there is a single point of intersection to prove tangency.

Warning: That tangency property only works for circles. It will not work for functions in general. (Though I think it might for a parabola, too. I'll have to think about it. It is best not to assume this.)

-Dan
• Dec 17th 2007, 03:20 AM
This post is wrong, sorry.
Quote:

Warning: That tangency property only works for circles. It will not work for functions in general. (Though I think it might for a parabola, too. I'll have to think about it. It is best not to assume this.)
This property does apply to parabolas, and any other function without a point of inflection.

The part about it working for parabolas is right though
• Dec 17th 2007, 03:50 AM