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
Best Answer
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 $$\alpha=\arctan12.$$