Problem on tangent to the parabola. Let PQ be a focal chord of the parabola $y^2= 4ax$. The tangents to the parabola at P and Q meet at a point lying on 
the line $y = 2x + a, a > 0.$ If chord PQ subtends an angle $\theta$  at the vertex of $y^2= 4ax$, then tan$\theta= ?$
My attempt 
Eqn of PQ $\rightarrow y=-2x+2a$ 
Solving it with parabola
$$y^2−4ax(\frac{2x+y}{2a})=0$$
$$⇒y^2−4x^2−2xy=0$$
For $x^2+y^2+2hxy=0$
$$\tan\theta=|2\sqrt{h^2-ab}/a+b|$$
using this formula, I got two value of tan$\theta$ one is positive and other is negative , which is correct negative or positive value ? 
 A: Tangents at the endpoints of a focal chord intersect on the directrix (line $x=-a$). Hence they intersect at $D=(-a,-a)$ and tangency points $P$ and $Q$ have coordinates
$$
y_{P/Q}=a(-1\pm\sqrt5),\quad x_{P/Q}={a\over2}(3\mp\sqrt5).
$$
Then, by the cosine rule:
$$
\cos\theta={OP^2+OQ^2-PQ^2\over2\,OP\cdot OQ}=-{3\over\sqrt{29}}
$$
and consequently
$$
\tan\theta=-{2\sqrt{5}\over3}.
$$
A: Focus: $(a,0)$ 
$$
2y\frac{dy}{dx}=4a\Rightarrow\frac{dy}{dx}=\pm\frac{2a}{\sqrt{4ax}}=\pm\sqrt{\frac{a}{x}}
$$
and 
$$
\frac{dy}{dx}=\frac{2a}{y}
$$
Let point $P$ have coordinates $(at^2,2at)$, then the equation of $PQ$ is 
$$
y=\frac{2at}{at^2-a}(x-a)=\frac{2t}{t^2-1}(x-a)
$$
To find the coordinates of $Q$ we need to solve
$$
\left(\frac{2t}{t^2-1}(x-a)\right)^2=4ax.
$$
We do not need to solve it directly, however. Simplifying,
$$
\left(\frac{2t}{t^2-1}\right)^2x^2-\left(2a\left(\frac{2t}{t^2-1}\right)^2 -4a\right)x+\left(\frac{2t}{t^2-1}\right)^2 a^2=0
$$
we know that the product of the two roots of the equation is $a^2$. So the coordinates of $Q$ can be found to be $(a/t^2,-2a/t)$.
Now note that 
$$
\frac{dy}{dx}=\frac{2a}{y}=\frac{2a}{2at}=1/t
$$
for point $P$ and 
$$
\frac{dy}{dx}=-t
$$
for point $Q$. When tangent at $P,Q$ intersects,
$$
\frac{1}{t}(x-at^2)+2at=-t(x-\frac{a}{t^2})-\frac{2a}{t}
$$
which simplifies to $x=-a$. At the point of intersection, $y=-a$, which can help us to find out that $t=\frac{1}{\sqrt2}$. (WOLOG we may ignore the negative value)
$\Rightarrow P(a/2,\sqrt2 a), Q(2a,-2\sqrt2 a)$.
$$
\tan(POQ)=\frac{\tan(POx)+\tan(QOx)}{1-\tan(POx)\tan(QOx)}=\frac{2\sqrt2+\sqrt2}{1-2\sqrt2\sqrt2}=-\sqrt2
$$
So $\theta$ is obtuse. Please tell me if there are some mistakes.
