Proof

[George Saliaris]

Let C1: (x-3)^2+y^2=5 C2: y^2=2x I) Show that the circle is tangent to the parabola (completed, contact points are A(2,2) and A'(2,-2)). II) Let M be a point of the parabola (different from (2,2),(2,-2)).From that point draw a tangent to the circle and let C be the contact point. Prove that (MC) =abs{ (ME) - 5/2},where E(1/2,0). What I thought/did is : M ε C2..etc., C ε C1, MC=MA or MA' (depends where M lies), some inner product of vectors. The result was writing a lot equations that did not get me somewhere. Is there a different/geometrical approach to this problem?

 

[Cubist]

Can you please explain your statement M.C = M.A ? (I think this might not be correct)

BTW: I would recommend that you put a "." in "M.A" if you intend dot product. Otherwise "MA" could be interpreted as the direction vector from point M to point A.

Have you been asked to solve the question with vectors?

 

[George Saliaris]

So, I looked carefully and saw that M.C is not equal to M.A. And no, I do not have to solve the problem with vectors.

 

[George Saliaris]

Here is a rough try of me trying to analyze what I have to prove. ('Γ' is the same as 'C')

 

[George Saliaris]

I have a strong feeling that in order to solve the question, I have to approach it geometrically.

 

[Cubist]

The focus of the parabola is at (0, 0.5) but I doubt that's useful. No, I can't think of a geometric way of doing this.

I personally think that algebra is the way forward. I'll have a go and post back if I get anywhere. But you should keep trying too. I'm thinking of finding a quadratic for the circle's intersect points (with a line passing through M with gradient m) and then letting the determinant=0 to get the two tangents

 

[George Saliaris]

Actually E(1/2,0) is the focus of the parabola. In the original problem it says that E is the focus of the parabola y^2=2x => 2p=1=>p=1/2=>E(p/2,0)=>E(1/2,0)

 

[Cubist]

I didn't make any progress with algebra, so I've started to think again about a geometric method. Here's an image to aid discussion...



E is the parabola focus. K is static, the circle centre.
M moves along the parabola, affecting the positions of both A and C

 

[Cubist]

OK I've solved it, it was fairly easy using the diagram above and Pythagoras (twice) in triangle MCK.

Hint: using the diagram write some (fairly) simple expressions for ME and MC in terms of the (x,y) coordinates of M. Then use the equation of the parabola to rewrite y in terms of x. Then simplify.

 

[George Saliaris]

It's abs(ME-5/2) but kk

 

[George Saliaris]

Omg thanks a lot, I got it. I could not
see the obvious.. -. Plus, the diagram I did was not helping me...

 

[Cubist]

I spent ages messing around with algebra until doing a decent diagram! To complete things...

E1) From the diag, ME = x + 0.5
E2) From the diag, using Pythagoras MC = sqrt( MK^2 - 5)
E3) From the diag, using Pythagoras MK^2 = (x-3)^2 + y^2

Using E2 & E3:
$$MC=\sqrt{\left(x-3\right)^2+y^2-5}$$
Use parabola eqn y^2 = 2*x

$$MC = \sqrt{\left(x-3\right)^2+2x-5}$$$$=\sqrt{x^2-4x+4}$$$$=\sqrt{\left(x-2\right)^2}$$$$=\left|x-2\right|$$
Using E1: $$MC=\left|ME-2.5\right|$$