Locus of intersection of two lines

Question

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

Answer 1

$$\begin{align} \triangle CDB \sim \triangle TAB \qquad\to\qquad \frac{y}{a+x} \;=\; \frac{t}{2a} \qquad\to\qquad \frac{t}{a} &\;=\; \frac{2y}{a+x} \\[6pt] \triangle CDA \sim \triangle OAT \qquad\to\qquad \frac{y}{a-x} \;=\; \frac{\;a\;}{t} \qquad\to\qquad \frac{t}{a} &\;=\; \frac{a-x}{y} \end{align}$$

$$ $$

$$\frac{2y}{a+x} = \frac{a-x}{y} \quad\to\quad x^2 + 2y^2 = a^2 \quad\to\quad \frac{x^2}{a^2} + \frac{y^2}{a^2/2} = 1$$ $$\to\quad \text{eccentricity }^2 \;=\; \frac{a^2 - a^2/2}{a^2} \;=\; \frac{1}{2} \qquad\square$$

Answer 2

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.