# Equation of tangent

1. Aug 17, 2010

### thereddevils

1. The problem statement, all variables and given/known data

Show that the tangent to the ellipse ax^2+by^2=1 at the point (h,k) has equation ahx+bky=1

Hence, deduce that the chord of contact of tangents from the point (m,n) to the ellipse ax^2+by^2=1 has equation amx+bny=1

2. Relevant equations

3. The attempt at a solution

I managed to prove the first part but am having problem with the second part.

Of course, it can be done by evaluating the gradient of tangent of the ellipse and use the straight line formulas to prove that.

But i am not sure how to DEDUCE that from what i got from the first part.

2. Aug 17, 2010

### HallsofIvy

Any line through (m,n) is of the form y- n= a(x- m) where "a" is the slope of the line. On the other hand, any tangent to the ellipse at (h, k) has equation ahx+ bky= 1 which we can rewrite as bky= -ahx+ 1 or y= -(ah/bk)x+ 1/bk which has slope -(ah/bk). A line that both passes through (m,n) and is tangent to the ellipse at (h, k) must be of the form y- n= -(ah/bk)(x- m) or, multiplying through by bk, bky- bkn= -ahx+ ahm or ahx+ bky= bkn+ ahm.

Now use the fact that (h, k) is a point on the ellipse: $ah^2+ bk^2= 1$

3. Aug 18, 2010

### thereddevils

thanks, this is the continuatino of the question.

Show that for all values of t, the chord of contact of tangents from the point (2t, 1-t) to the ellipse ax^2+by^2=1 passes through a fixed point and determine the point of this coordinates.

The cartesian equation of (2t,1-t) is y=-1/2 x+1

So any point from this line to the ellipse will pass through a fixed point say (p,q)

Any further hints on this?