Proving that locus represents an ellipse with eccentricity 1/√2

Aug 2019
88
23
India
The question is
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 the diameter through A; prove that the locus of the intersection of AP and BT is an ellipse whose eccentricity is \(\displaystyle \frac{1}{\sqrt2}\) .

I have made this picture for this problem is :

Screen Shot 2019-12-25 at 1.36.03 PM.png




\(\displaystyle A = (a_1,a_2)\), \(\displaystyle P=(x_1,y_1)\), equation to circle can be considered for simplicity as \(\displaystyle x^2 + y^2 = r^2 \implies x^2+y^2 = a_1^2 + a_2^2 \). We can see very clearly (can even prove easily) that the coordinates of \(\displaystyle B\) is \(\displaystyle (-a_1,-a_2)\).

The theorem on chord of contact says If from T=(x',y') two tangents are drawn to the circle \(\displaystyle x^2 + y^2 = a^2\) then the equation to the chord of contact is \(\displaystyle x'x +y'y = a^2~~~~~~~(1)\) . In our figure the chord of contact is AP, the equation of AP is \(\displaystyle \frac{y-a_2}{x-a_1} = \frac{y_1 - a_2}{x_1 - a_1} \), doing some rearrangements we have :
\(\displaystyle (a_2 - y_1)x + (x_1 - a_2)y = a_2x_1 - a_1y_1 \). Now, comparing this equation of chord of contact AP with our equation (1) we find the coordinates of T (our own T, the one in figure) are \(\displaystyle (a_2-y_1, x_1-a_2)\).

Let the coordinates of C be \(\displaystyle (c_x, c_y)\). Slope of line BT is same no matter if we calculate it using B and T or B and C, therefore
\(\displaystyle \frac{c_y + a_2}{c_x+a_1} = \frac{x_1-a_1+a_2}{a_2-y_1+a_1}~~~~~~~~~~~~~~(i) \).

Same argument for AC and AP, we have
\(\displaystyle \frac{c_y - a_2}{c_x - a_1} = \frac{y_1 - a_2}{x_1 - a_1} ~~~~~~~~~~~~~~(ii) \)

When we want to find the locus we usually find the relation between x and y coordinates in terms of known things. In our case, we have to find the relation between \(\displaystyle c_x\) and \(\displaystyle c_y\) independent of \(\displaystyle x_1\) and \(\displaystyle y_1\). So, what I did is to solve the equations (i) and (ii) for x1 by writing \(\displaystyle y_1 = \sqrt{a_1^2 + a_2^2 - x_1^2} \) but doing it is almost impossible by hand and doing it on computer results in very bad value for x1.

Where am I wrong in my approach? Why I'm not getting the correct thing? Are my reasonings faulty?

Please help. Any suggestion will be much appreciated.

Thank you.