Chord of Contact Formula Proof (1 Viewer)

[someth1ng]

Can someone explain it to me? It's the weirdest proof...ever.

 

[barbernator]

http://www.slideshare.net/nsimmons/x2-t03-06-chord-of-contact-and-properties-2010

 

[deswa1]

Its an elegant proof. What you need to be careful of is confusing fixed point (x1,y1) etc. with the general points (x,y)

 

[Nooblet94]

 

[Drongoski]

Its an elegant proof. What you need to be careful of is confusing fixed point (x1,y1) etc. with the general points (x,y)

Same approach as for chord of contact for parabola & hyperbola. As the Cambridge book points out, it is a really elegant proof. And the beauty is it looks exactly like the eqn of the tangent, except the point is not on the ellipse.

 

[funnytomato]

View attachment 24895

I just uploaded it

 

[funnytomato]

@OP, the most important thing is probably the last statement in the post above

 

[funnytomato]

Same approach as for chord of contact for parabola & hyperbola. As the Cambridge book points out, it is a really elegant proof.

elegant = weird

 

[Nooblet94]

elegant = weird

Hardly.

 

[funnytomato]

Hardly.

for this proof, as least when I first saw it

 

[someth1ng]

In the tangent formulae to chord of contact formulae, isn't it just that the co-ordinates P(x1, y1) and Q(x2,y2) are represented as x and since you know that both can be subbed in to the straight line and be true, then it must be true?

 

[funnytomato]

In the tangent formulae to chord of contact formulae, isn't it just that the co-ordinates P(x1, y1) and Q(x2,y2) are represented as x and since you know that both can be subbed in to the straight line and be true, then it must be true?

yeah, since it can shown the line xx0/a^2+yy0/b^2=1 (where x0 and y0 are constants) pass through both P (since x1x0/a^2+y1y0/b^2=1) and Q (since x2x0/a^2+y2y0/b^2=1)

and two points determine a unique line