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
 A: If $y=mx+n$ is a tangent then $$n^2=a^2m^2+b^2$$ or
$$n^2=3m^2+\frac{3}{2}.$$
Also, we have $2=m+n$ and we got the following equation on slopes:
$$(2-m)^2=3m^2+\frac{3}{2}.$$
After this use $$\tan\alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.$$
I got $$\alpha=\arctan12.$$
A: Let $m$ be the slope of a tangent to the ellipse from $(1,2)$.
The equation of the tangent is $y=mx-m+2$.
Put it into the equation of the ellipse,
\begin{align*}
x^2+2(mx-m+2)^2&=3\\
(1+2m^2)x^2-4m(m-2)x+2m^2-8m+5&=0
\end{align*}
As the tangent meets the ellipse at only one point.
\begin{align*}
[4m(m-2)]^2-4(1+2m^2)(2m^2-8m+5)&=0\\
(4m^4-16m^3+16m^2)-(4m^4-16m^3+12m^2-8m+5)&=0\\
4m^2+8m-5&=0
\end{align*}
If $m_1$ and $m_2$ are the slopes of two tangents, then $\displaystyle m_1+m_2=-2$ and $\displaystyle m_1m_2=-\frac{5}{4}$.
Let $\theta$ be the acute angle between the two tangents.
\begin{align*}
\tan\theta&=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\\
\tan^2\theta&=\frac{(m_1+m_2)^2-4m_1m_2}{(1+m_1m_2)^2}\\
&=\frac{(-2)^2-4(\frac{-5}{4})}{(1+\frac{-5}{4})^2}\\
&=144\\
\theta&=\tan^{-1}(12)
\end{align*}
A: 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 $cos(\theta-\alpha)=\sin \alpha\sin \theta+\cos \alpha \cos \theta=\frac 3{\sqrt {a^2+16 b^2}}$
