Show that the angle between $OP$ and the normal to the curve at $P$ satisfies the following I'm struggling to answer the following question below

I've already worked out the gradient to the curve at $P$, but I'm having difficulty answering the second part of the question. MY attempt is as follows:
I focus on $\triangle OPQ$, where $Q$ is the intersection of the normal at $P$ to the x-axis
Re-writing $P$ as $(x', y')$ to avoid confusion.
The angle $\theta_2$ between $OP$ and the x-axis satisfies $\tan(\theta_2)=\frac{y'}{x'}$
The equation of the normal to the curve at $P$ is $(y-y')=(\frac{ax+y}{x+ay})(x-x')$, substituting $y=0$ into the equation gives the x co-ordinate of $Q$, which gives $x=(x'(\frac{ax+y}{x+ay})-y')(\frac{x+ay}{ax+y})=x'-y'(\frac{x+ay}{ax+y})$
So the length of $OQ$ is $x'-y'(\frac{x+ay}{ax+y})$
The length of $OP$ is $\sqrt{x'^2+y'^2}$
And $\angle QOP$ satisfies $tan(\theta_2)=\frac{y'}{x'}$
But by using the cosine rule to get $\angle OPQ$, I get nothing close to the desired result. Any hints or different approaches would be greatly appreciated.

Notice, gradient of the curve: $x^2+y^2+2axy=1$  $$\frac{d}{dx}(x^2+y^2+2axy)=\frac{d}{dx}(1)$$ $$2x+2y\frac{dy}{dx}+2ax\frac{dy}{dx}+2ay=0$$ $$2(ax+y)\frac{dy}{dx}=-2(x+ay)$$ $$\frac{dy}{dx}=\frac{-2(x+ay)}{2(ax+y)}$$  $$\implies \color{blue}{\text{slope of tangent}\,, \frac{dy}{dx}=\frac{-(x+ay)}{ax+y}}$$
Above is the gradient of the curve (i.e. gradient of tangent) at the point $(x, y)$.  
Now, the slope of line OP joining origin $O(0, 0)$ & $P(x, y)$ $$=\frac{y-0}{x-0}=\frac{y}{x}$$ & the slope of normal at $P(x, y)$ $$=\frac{-1}{\text{slope of tangent at P}}=\frac{-1}{\frac{-(x+ay)}{ax+y}}=\frac{ax+y}{x+ay}$$ Hence, the angle $\theta$ between OP, slope, $m_1=\frac{y}{x}$  & normal at P, slope, $m_2=\frac{ax+y}{x+ay}$ is given by the following formula $$\tan \theta=\frac{|m_1-m_2|}{1+m_1m_2}$$ $$\implies \tan \theta=\frac{\left|\frac{y}{x}-\frac{ax+y}{x+ay}\right|}{1+\frac{y}{x}\times \frac{ax+y}{x+ay} }$$ $$=\frac{\left|ay^2-ax^2\right|}{x^2+axy+axy+y^2}$$ $$\tan \theta=\frac{a\left|y^2-x^2\right|}{x^2+y^2+2axy}\quad \text{since},\ 0<a<1$$ Setting the value of $x^2+y^2+2axy$, we get $$\tan \theta=\frac{a\left|y^2-x^2\right|}{1}$$
$$\implies \color{blue}{\tan \theta=a\left|y^2-x^2\right|}$$
Hence, $\theta$ is duly satisfying above relation
