<s>[INST] Locus of intersection of two lines If the tangent at any point P of a circle $x^2 + y^2 = a^2$  meets the tangent at a fixed point A $(a,0)$ in T and T is joined to B , the other end of the diameter through A . Then we have to prove that the locus of intersection of AP and BT is an ellipse whose eccentricity is $\frac{1}{\sqrt{2}}$
I tried ,
Let P$ (h,k)$ on the circle . 
The equation of tangent is    $hx + ky = a^2 $     and through A it is $x = a$
So the point is $(a , \frac{a^2 - ha}{ k })$
Equation of AP is $( h-a)y = k(x-a)$ 
Equation of BT is$ ky = (a-h ) ( x+a) $
Then I tried to solve both the equation , but I am not getting the result . 
Thank you in advance
 [/INST] 
$$\begin{align}
\triangle CDB \sim \triangle TAB \qquad\to\qquad \frac{y}{a+x} \;=\; \frac{t}{2a} \qquad\to\qquad \frac{t}{a} &\;=\; \frac{2y}{a+x} \\[6pt]
\triangle CDA \sim \triangle OAT \qquad\to\qquad \frac{y}{a-x} \;=\; \frac{\;a\;}{t} \qquad\to\qquad \frac{t}{a} &\;=\; \frac{a-x}{y}
\end{align}$$
$$ $$
$$\frac{2y}{a+x} = \frac{a-x}{y} \quad\to\quad x^2 + 2y^2 = a^2 \quad\to\quad \frac{x^2}{a^2} + \frac{y^2}{a^2/2} = 1$$
$$\to\quad \text{eccentricity }^2 \;=\; \frac{a^2 - a^2/2}{a^2} \;=\; \frac{1}{2} \qquad\square$$
</s>