Find the locus of points $P$ for which angle bisector of tangents to a standard ellipse is $y = x \tan \theta$. 
Find the locus of points $P$ for which angle bisector of tangents to a standard ellipse is $y = x \tan \theta$.

Let $ P = (u, v)$ and ellipse be $E = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1$
Then joint tangent equation for the point $P$ is $$J(x,y) = \dfrac{(y-v)^2}{b^2} + \dfrac{(x-u)^2}{a^2} - \dfrac{(uy-vx)^2}{a^2b^2} $$
I have to find $(u,v)$ for which the angle bisectors of line pair $J$ are $ y - x\tan \theta = 0$ and $y + x\cot \theta -v - u \cot \theta = 0$.
Translating $P$ - the centre of $J$ - to origin with $x = X + u$ and $y = Y + v$:
$$J'(X, Y) = J(X+u, Y+u) = X^2 \left (\dfrac1{a^2} - \dfrac{v^2}{a^2b^2}\right) 
 + Y^2 \left (\dfrac1{b^2} - \dfrac{u^2}{a^2b^2}\right) + \dfrac{2XYuv}{a^2 b^2}$$
The translated angle bisectors are $Y - X\tan \theta + v - u \tan \theta = 0$ and $Y + X \cot \theta = 0$.
The angle bisectors of $J'$ is given by $$X^2 \dfrac{uv}{a^2b^2} + \left(\dfrac1 {b^2} -\dfrac{u^2}{a^2b^2} -\dfrac{1}{a^2} +\dfrac{v^2}{a^2b^2} \right)XY - \dfrac{uv}{a^2b^2} Y^2 = 0\tag 1$$
or $$(Y + X \cot \theta) (Y - \tan \theta X + v - u \tan \theta) = 0\tag 2$$
Equating corresponding terms of $(1)$ and $(2)$ gives $$\begin{cases}3) : uv = -a^2b^2 \\4) : v = u \tan \theta \\5) : \dfrac1 {b^2} -\dfrac{u^2}{a^2b^2} -\dfrac{1}{a^2} +\dfrac{v^2}{a^2b^2} = \cot \theta - \tan \theta \end{cases}$$
From $(3)$ and $(5)$ I get $u^2 - v^2 -uv(\cot \theta - \tan \theta) = a^2 - b^2$

The answer given is $u^2 - v^2 -2uv\cot \theta = a^2 - b^2$.
Where did I go wrong ? 
Is there any alternative way to do this questions, my method is very tedious ?
 A: I think there is something wrong in this question, unless I'm missing something. Requesting that the angle bisector of the tangents issued from $P$ is $y=x\tan\theta$ is the same as requesting that the angle bisector passes through $O=(0,0)$, that is through the center of the ellipse (assuming that a "standard ellipse" has its center there).
If $A$ and $B$ are the tangency points and $M$ their midpoint, it is a property of the ellipse that line $PM$ passes through $O$. Hence, to satisfy the request, the bisector of $\angle APB$ must pass through $M$, which entails $PA=PB$. But this condition is satisfied only if $P$ belongs to the axes of the ellipse (unless the ellipse is a circle).
In addition, I don't understand the role of $\theta$: if it is given (as it seems to be), then $P$ must belong to the given line $y=x\tan\theta$, what is the meaning of asking for the locus of $P$?
EDIT.
Your solution is right, but substituting there $\tan\theta=v/u$ (from eq. $(4)$) you get $0=a^2-b^2$. The only way to have a solution is then if $a=b$ (the ellipse is a circle) or if either $\tan\theta$ or $\cot\theta$ doesn't exist ($P$ lies on the coordinate axes).
