Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$. 
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.

The given ellipse is $\dfrac{x^2}{3}+\dfrac{y^2}{\frac{3}{2}}=1$
Any point on the ellipse is given by $(a\cos \theta,b\sin \theta)$ where $a=\sqrt 3,b=\frac{\sqrt 3}{\sqrt 2}$.
Now  slope of  the tangent to the curve at  $(a\cos \theta,b\sin \theta)$ is $\dfrac{-a\cos \theta}{2b\sin \theta}$.
Hence we have $\dfrac{b\sin \theta- 2}{a\cos \theta -1}=\dfrac{-a\cos \theta}{2b\sin \theta}$.
On simplifying we get $4b\sin \theta +a\cos\theta =3$
If we  can find the value of $\theta $ from above then we can find the two points on the ellipse where the tangents touch them but I am unable to solve them.
Please help to solve it.
Any hints will be helpful

Let $\alpha$ be the angle for which $\cos\alpha = \frac a{\sqrt{a^2+16 b^2}}$ and $\sin \alpha=\frac {4b}{\sqrt{a^2+16 b^2}}$.
Then $\theta-\alpha$ can be solved from $cos(\theta-\alpha)=\sin \alpha\sin \theta+\cos \alpha \cos \theta=\frac 3{\sqrt {a^2+16 b^2}}$
