Parabola

**slevvio** · February 28th 2008, 06:42 AM

Let a,b be positive numbers. Sketch the parabolas

$y^2=4a^2-4ax$
$y^2=4b^2+4bx$

and show that they have a common focus, and that they intersect at right angles.

I was able to show that the focus of each parabola was at the origin by reducing these equations to standard form:

$y^2=-4a(x-a)$
$y^2=4b(x+b)$

But I don't know how to prove that they intersect at right angles. I got the coordinates $\left( \frac{a^2-b^2}{a+b},\pm 2\sqrt{ab} \right)$ as the points of intersection. How do I show that the parabolas cross through these points at right angles to one another? Any help would be appreciated

.

**slevvio** · February 28th 2008, 06:54 AM

If the parabolas intersect at right angles, is it safe to say that the tangents at those points are perpendicular ? i.e. the gradients are negative reciprocals of one another?

**Mathnasium** · February 28th 2008, 01:03 PM

Yes, it's safe to say that the tangents are also perpendicular, so I think you've got it - find the slope of the tangent line to each of the two parabolas at the points of intersection by differentiating the equation for each parabola. The slopes of the tangents should be negative reciprocals.

(Side note: Can't you simplify the x-coordinate of the points of intersection by factoring the numerator as a difference of two squares and then canceling one of the terms with the denominator?)

It looks like you'll have to either do implicit differentiation or solve for y by taking the square root of both sides.

Implicitly, for the first one, you get:

$2yy' = -4a$ (since a and b are constants), so:

$y' = \frac{-4a}{2y} = \frac{-2a}{y}$ .

For the second one, you get $2yy' = 4b$ , so:

$y' = \frac{4b}{2y} = \frac{2b}{y}$ .

As the OP pointed out below, these are negative reciprocals of one another, verifiable by cross-multiplication, when $y = 2\sqrt{ab}$ .

Two brains are better than one.

**slevvio** · February 28th 2008, 01:46 PM

Your differentiation isn't quite right - there is no term in x in the result.

$y' = -\frac{2a}{y}$

If you carry out implicit differentiation for both parabolas, then put the value of $y=2\sqrt{ab}$ into the derivative, the values of the tangents are negative reciprocals of one another. Thanks for your help!

**Mathnasium** · February 29th 2008, 01:05 AM

Thanks - I'll edit my original post. Too busy learning LATEX, unlearning calculus, I guess. Only so much room in my brain.

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