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
 A: Finding the point of intersection
\begin{alignat*}{2}
AT{:}&\quad & x - a &= 0 \\
PT{:}&\quad & hx + ky - a^2 &= 0 \\
T{:}&\quad & \left(a,\frac{a(a-h)}{k}\right) \\
AP{:}&\quad & kx + (a-h)y - ak &= 0 \\
BT{:}&\quad & (h-a)x + 2ky + a(h-a) &= 0
\end{alignat*}
Then the point of intersection between these is
$$
Q:
\frac a{(a-h)^2 + 2k^2}
\Bigl(
2k^2-(a-h)^2,
2k(a-h)
\Bigr)
$$
I computed all of the above using homogeneous coordinates and cross products. If you have two points $(x_1,y_1)$ and $(x_2,y_2)$, then the vector $(a,b,c)=(x_1,y_1,1)\times(x_2,y_2,1)$ represents the line $ax+by+c=0$ joining these. If you have two lines $a_1x+b_1y+c=0$ and $a_2x+b_2y+c_2$ then the cross product $(x,y,z)=(a_1,b_1,c_1)\times(a_2,b_2,c_2)$ represents the point of intersection $(x/z,y/z)$.
Attempt to eliminate one variable
Now it is important to realize that $P$ is not just any point, but a point on the circle. So $h^2+k^2=a^2$. You can write all occurrences of $k^2$ as $a^2-h^2$. Unfortunately there is an unsquared $k$ in the second coordinate which you can write as $k=\pm\sqrt{a^2-h^2}$.
$$
Q:
\frac a{(a-h)^2 + 2(a^2-h^2)}
\Bigl(
2(a^2-h^2)-(a-h)^2,
\pm2\sqrt{a^2-h^2}(a-h)
\Bigr)
$$
Perhaps you want to continue from here yourself.
Avoiding one variable
Personally, I don't like square roots. Which means I'd probably start over, writing $P$ using the tangent half-angle substitution as
$$P: \frac a{t^2+1}\Bigl(t^2-1,2t\Bigr)$$
For $t\in\mathbb R$ this covers all points on the circle except for $A=(a,0)$. Doing the same process outlined above, you will now end up with coordinates in a single parameter (namely $t$) instead of two parameters ($h$ and $k$).
\begin{alignat*}{2}
AT{:}&\quad & x - a &= 0 \\
PT{:}&\quad & (t^2-1)x + 2ty - a(t^2+1) &= 0 \\
T{:}&\quad & \left(a,\frac at\right) \\
AP{:}&\quad & tx + y - at &= 0 \\
BT{:}&\quad & x - 2ty + a &= 0 \\
Q{:}&\quad & \frac{a}{2t^2+1}\left(2t^2-1,2t\right) \\
\end{alignat*}
Spotting the ellipse
This looks again very similar to the tangent half angle substitution. Writing $u$ for $\sqrt2t$ we have
$$Q = \frac{a}{u^2+1}\left(u^2-1,\sqrt2u\right)$$
If you scale the $y$ coordinate by $\sqrt2$ you are back at a point on the circle of radius $a$ around the origin (simply compare it to the formula for $P$). Which means that $Q$ must lie on an ellipse which is that circle scaled by $1/\sqrt2$ in the $y$ direction. Its semimajor axis is $a$, its semiminor axis $b=a/\sqrt2$.
Implicit form
If (in a similar problem) you have problems spotting the ellipse, it might be useful to turn the equation into its implicit form:
\begin{align*}
(2t^2+1)x &= (2t^2-1)a &
(2t^2+1)y &= 2ta \\
2(x-a)t^2 + (x+a) &= 0 &
2yt^2 + 2at + y &= 0
\end{align*}
So you are looking for the common solutions of two equations which are quadratic in $t$. You can use the resultant to eliminate $t$. Computing it as the determinant of the Sylvester matrix you get
$$
\begin{vmatrix}
2(x-a) & 0 & (x+a) & 0 \\
0 & 2(x-a) & 0 & (x+a) \\
2y & 2a & y & 0 \\
0 & 2y & 2a & y
\end{vmatrix}
= 8a^2(x^2+2y^2-a^2)
$$
So the ellipse is described by
$$x^2+2y^2-a^2=0$$
For $y=0$ you get $x=\pm a$ so its semimajor axis is $a$. Conversely for $x=0$ you get $y=\pm\sqrt{1/2}a$ so its semiminor axis is $b=\sqrt{1/2}a$. So the result here matches the one from the previous section.
Computing the eccentricity
Following Wikipedia the eccentricity is now
$$e=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{a^2-a^2/2}{a^2}}=\sqrt{\frac12}$$
as claimed in the problem statement.
A: 
$$\begin{align}
\triangle CDB \sim \triangle TAB \qquad\to\qquad \frac{y}{a+x} \;=\; \frac{t}{2a} \qquad\to\qquad \frac{t}{a} &\;=\; \frac{2y}{a+x} \\[6pt]
\triangle CDA \sim \triangle OAT \qquad\to\qquad \frac{y}{a-x} \;=\; \frac{\;a\;}{t} \qquad\to\qquad \frac{t}{a} &\;=\; \frac{a-x}{y}
\end{align}$$
$$ $$
$$\frac{2y}{a+x} = \frac{a-x}{y} \quad\to\quad x^2 + 2y^2 = a^2 \quad\to\quad \frac{x^2}{a^2} + \frac{y^2}{a^2/2} = 1$$
$$\to\quad \text{eccentricity }^2 \;=\; \frac{a^2 - a^2/2}{a^2} \;=\; \frac{1}{2} \qquad\square$$
