Solve Equation of Tangent to an Ellipse at Point P

[Firepanda]

Sorry title was supposed to be Conic Sections, but my I key is sticky :)

I had a question today, It went somethng like this:

An epllipse of equation ((x^2)/4) + y^2 = 1

Find the equation of the tangent which passes through point P: (4,0)

Well this was a mock exam question, where no answers were available. I keep stumbing on questions of this format and I can never do them.

There is bound to be a routine to go through when caculating something like this and it would be great to know it.

So far I have differentiate implicitly to get dy/dx = -x/4y, point P isn't on the curve, so I have to find (x,y) point on the curve that joins it and P as a tangent.

Any help? Thanks

 

[HallsofIvy]

First, write the point as (x₀, y₀) to distinguish from a general (x, y) point. Yes, the slope of a tangent line to the ellipse at that point is -x₀/4y₀ and so the equation of the tangent line is y= (-x₀/4y₀)(x- x₀)+ y₀. In order that that line go through (4, 0), you must have 0= (-x₀/4y₀)(4- x₀)+ y₀.

In order that P be on the ellipse it must also be true that x₀²/2+ y₀²= 1. That gives you two equations to solve for x₀ and y₀.

 

[Firepanda]

I understood your 1st paragraph, but not so much the last sentence.

When you say in order that P be on the ellipse, I don't want it to be on the ellipse, I want it to be a point the tangent is connected to.

http://img142.imageshack.us/img142/6224/ellipsecm5.jpg

 

[Firepanda]

Nevermind I got it :D Thanks!