Finding the point through which the tangent passes 
A hyperbola passes through the point $P=(\sqrt{2},\sqrt{3})$  and has foci $(\pm 2,0)$. Show that the tangent at $P$ passes through the point $(2\sqrt{2},3\sqrt{3})$.

Attempt:
So the equation of hyperbola is 
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
We have been given $ae=2$ and a point on hyperbola.
So it gives us two equations:  
$$a^2+b^2=4\tag{1}$$
$$2b^2-3a^2=(ab)^2\tag{2}$$ 
Substituting $b^2=4-a^2$ and solving with $(2)$, I get the equation $a^4-9a^2+8=0$. This gives me the value of $a$ to be $\{1,-1,2\sqrt{2},-2\sqrt{2}\}$. None of this satisfies the condition of hyperbola... What did I do wrong?
 A: There’s a property of hyperbolas that leads to a simple way of solving this problem instead of trying to find an equation of the hyperbola: the tangent to a hyperbola bisects the angle formed by the lines to its foci.  
The lines through $\left(\sqrt2,\sqrt3\right)$ and the two foci have direction vectors $\mathbf v=\left(\sqrt2-2,\sqrt3\right)$ and $\mathbf w=\left(\sqrt2+2,\sqrt3\right)$. The relevant angle bisector has direction vector $\|\mathbf w\|\mathbf v+\|\mathbf v\|\mathbf w = \left(4,4\sqrt6\right)$, i.e., the slope of the tangent at $P$ is $\sqrt6$. With this slope in hand, verifying that $\left(2\sqrt2,3\sqrt3\right)$ lies on this line should be a simple matter.  
I’m not sure why you think that you’ve done anything wrong in your attempt at a solution, though. You just haven’t carried it all the way through. You can discard negative values of $a$ right off the bat. If you substitute $a=2\sqrt2$ into the first equation, you get $8+b^2=4$, which has no real solutions, so that value can also be discarded. For $a=1$, $b=\pm\sqrt3$, and of course we discard the negative solution again.
