Find the equation of  tangents to the ellipse x² + 2y² = 3 drawn from the point (1, 2) also find the angle between these tangents.

Given ellipse is $x^2+2y^2=3 and point=(1,2)$

The equation of pair of tangents to the ellipse is $S_1^2=S.S_{11}$

${\left({\mathrm{xx}}_1+2{\mathrm{yy}}_1-3\right)}^2=\left({\mathrm{x}}^2+2{\mathrm{y}}^2-3\right)\left({\mathrm{x}}_1^2+2{\mathrm{y}}_1^2-3\right)$
At point (1, 2)
$\begin{array}{l}(\mathrm{x}+4\mathrm{y}-3{)}^2=\left({\mathrm{x}}^2+2{\mathrm{y}}^2-3\right)(1+8-3)\\ {\mathrm{x}}^2+16{\mathrm{y}}^2+9+8\mathrm{xy}-24\mathrm{y}-6\mathrm{x}=6{\mathrm{x}}^2+12{\mathrm{y}}^2-8\\ 5{\mathrm{x}}^2-8\mathrm{xy}-4{\mathrm{y}}^2+6\mathrm{x}+24\mathrm{y}-27=0\end{array}$
Represents pairs of tangents

$If \theta is angle between these \tan gents then$
$\begin{array}{l}\tan \mathrm{\theta }=\frac{2\sqrt{{\mathrm{h}}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}=\frac{2\sqrt{16-5(-4)}}{1}=2\times 6=12\\ \mathrm{\theta }={\tan }^{-1}\left(12\right)\end{array}$