# Homework Help: Find the eccentricity of this ellipse

1. Dec 12, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
The tangent at any point P of a circle meets the tangent at a fixed point A in T, and T is joined to B, the other end of diameter through A. Prove that the locus of point of intersection of AP and BT is an ellipse whose eccentricity is $1/ \sqrt{2}$

2. Relevant equations

3. The attempt at a solution
The very first thing I do is assume the equation of a circle. The next thing is to write the equations for tangents and solve them to get T. But it is getting complicated as nothing is known to me. So there are a number of variables which can't be eliminated. Any other ideas?

2. Dec 13, 2012

### tiny-tim

hi utkarshakash!
show us what you've tried, and where you're stuck, and then we'll know how to help!

alternatively, since the eccentricity is 1/√2, the minor axis must be 1/√2 time the major axis …

so have you tried squashing the whole diagram by 1/√2 (along AB), so that the the final result is a circle?

3. Dec 14, 2012

### utkarshakash

For the sake of simplicity, let the equation of the circle be $x^2 + y^2 = c^2$.

Let A = (x1,y1) and P = (x2,y2)

Equation of tangent at P
$xx_2 + yy_2 - c^2 = 0$
Equation of tangent at A
$xx_1 + yy_1 -c^2 = 0$

When I solve these two equations I get

$x = \dfrac{c^2 (y_2 - y_1)}{x_1y_2-x_2y_1} \\ y = \dfrac{c^2 (x_1 - x_2)}{x_1y_2-x_2y_1}$

OMG It looks so dangerous! I don't want to proceed ahead as the calculation will be complex and rigorous.

4. Dec 14, 2012

### tiny-tim

no!

A is fixed, so you can simplify by letting A = (c,0)

(and P = (ccosθ,csinθ) )

Last edited: Dec 14, 2012
5. Dec 15, 2012

### utkarshakash

That did simplify the expression to a large extent. Thanks!