parabola's tangents form a right angle

[music]

Many years ago, in college, a textbook ordered me to prove that from any point on the directrix, the two tangents that can be drawn to the parabola intersect at a right angle. (Maybe I've got that wrong, but that's how I remember the problem.)

 

I worked on that for months and was never able to solve it, even after the professor told me how to do it. Drove me nuts. So I've never forgotten it.

 

But I've also never found a proof of it. I'd appreciate seeing it proven if anyone cares. But of course I have no right to take anyone's time, so... no obligation or anything. Just something I've been curious about for a long time.

[doG]

See http://www.geometryatlas.com/entries/352

[music]

See http://www.geometrya...com/entries/352

 

Thank you!

 

If that is a proof, I haven't understood it yet, but I'll look at it for awhile.

[studiot]

It is more usual to prove that the locus of points where the tangents are perpendicular is the directrix to avoid solving a quadratic and is an equivalent question, (this was Dog's proof) but here goes one proof.

 

Let the parabola be represented in standard form

 

 

[math]{y^2} = 4ax[/math]

 

Then the directrix is the line

 

 

[math]x = - a[/math]

 

and the tangent is the line

 

 

[math]y = mx + \frac{a}{m}[/math]

 

Let the point of intersection of any two such tangents be the point

 

 

[math]{x_1}{y_1}[/math]

 

Then the equation of any tangent through this point is

 

 

[math]{y_1} = m{x_1} + \frac{a}{m}[/math]

 

Rarranging we see that this is a quadratic in the slope, m

 

 

[math]{x_1}{m^2} - {y_1}m + a = 0[/math]

 

But we are told this lies on the directrix as well so satisfies

 

 

[math]{x_1}=-a[/math]

 

 

therefore substituting

 

 

[math] - a{m^2} - {y_1}m + a = 0[/math]

 

From the standard properties of quadratic equations

 

 

[math]{\rm{product}}\,{\rm{of}}\,{\rm{roots = }}\frac{{{\rm{constant}}\,{\rm{term}}}}{{{\rm{coefficientof}}\,{{\rm{x}}^{\rm{2}}}}}[/math]

 

 

So this quadratic has two solutions

 

 

[math]{m_1}{m_2} = \frac{a}{{ - a}} = - 1[/math]

 

Which is the condition for the two lines to be perpendicular

Edited June 5, 2012 by studiot

[studiot]

music, was there something wrong with my last post?

[music]

studiot, your proof is beautiful! I don't know that "product of roots" rule so I can't follow that point, but the proof strikes me as both simple and clever.

[studiot]

 

Yes as well as not tesching much (if any) coordinate geometry these days they don't teach properties of quadratic equations.

 

There are two the other one being

 

 

[math]{\rm{sum}}\,{\rm{of}}\,{\rm{roots = }}\frac{{{\rm{coefficient}}\,{\rm{of}}\,{\rm{x}}}}{{{\rm{coefficient}}\,{\rm{of}}\,{{\rm{x}}^{\rm{2}}}}}[/math]

 

 

At one time, beofe the formula became popular, this used to be the way solve quadratic equations. The above two properties are still useful in agebraic manipulation as here, however.

 

go well

[imatfaal]

Surely there must be a minus sign in there somewhere

[studiot]

Yes well spotted there should be a minus sign.

 

I blame the arthritis in the typing finger.

[music]

I hadn't seen it before, but it's easy to understand: http://regentsprep.org/Regents/math/algtrig/ATE4/natureofroots.htm