# Prove that a variable chord of ellipse

1. Dec 11, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
Prove that a variable chord of ellipse which subtends 90° at the centre is always tangent to a concentric circle

2. Relevant equations

3. The attempt at a solution
I assume the simplest equation of ellipse to be
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
and the variable chord to be
y=mx+c

Now homogenising the given line with the equation of ellipse

$b^2x^2+a^2y^2-a^2b^2 \left( \dfrac{y-mx}{c} \right) ^2 = 0 \\ a^2b^2(1+m^2)=c^2(a^2+b^2)$

Now I assume the simplest equation of circle to be $x^2 + y^2 = k^2$
I have to prove that y=mx+c is a tangent to given circle
Applying condition of tangency I have to prove that
$k = \dfrac{c}{\sqrt{1+m^2}}$

In the RHS I substitute the value of (1+m^2) derived earlier. This way the RHS term becomes

$\dfrac{ab}{\sqrt{a^2 + b^2}}$

But the LHS is k. I'm confused here!

2. Dec 11, 2012

### Vargo

If you pick a point O, then every line m that does not pass through O is tangent to some circle with center O. To see this, drop a perpendicular segment from O to the line m. Then use that segment as the radius of the circle.

Now take O to be the center of your ellipse and take the chord to be the one given. The chord extends to a line, so there is a concentric circle tangent to that line. But the question is whether the point of tangency lies on the original chord or whether it lies on the extended line somewhere outside the ellipse. Hint: XOY is a right triangle. Forget the ellipse and focus on the fact that you are trying to determine whether the perpendicular dropped from O to line XY lies between X and Y.

3. Dec 11, 2012

### utkarshakash

The point of contact always lies on the chord and not on the extended line. Also what is wrong in my solution?

4. Dec 11, 2012

### haruspex

Not sure what's bothering you. You only have to show there is some concentric circle it's tangent to. You have successfully calculated its radius.

5. Dec 12, 2012

### utkarshakash

Yes I later realized that I actually arrived at the answer because the RHS as well as LHS both are constant. This means the radius of the circle is the one calculated above.

6. Dec 12, 2012

### Vargo

Would you mind explaining how you got that equation between a,b,m,c? I didn't understand that step. Does it use the fact that the chord subtends a right angle with the center? I can't see how you got there, but it looks like a neat trick.

7. Dec 12, 2012

### utkarshakash

Yes Of course. See my question. It says that the chord subtends a right angle at the centre. So I used the concept of homogenisation to arrive at that equation.

8. Dec 12, 2012

### Vargo

I see. You homogenize the two equations to get a quadratic form whose zero set is a pair of orthogonal lines. Therefore the trace of the form is zero giving you the equation. Cool trick.

I am still not clear on your proof though. You have a nice calculation of the radius of the tangent circle in terms of the parameters of the ellipse a and b. But have you proved that the point of contact is on the chord rather than the extended line y=mx+b? Somehow, you must address this explicitly because that is the crux of what you are being asked to prove. (Otherwise the subtended angle would be irrelevant and the statement would be trivial).

9. Dec 12, 2012

### utkarshakash

You can try out by drawing the figures. You will see that the point of contact always lies on the chord and not on the extended line if the circle is concentric.