If foci of an ellipse be (-1, 2) and (-2, 3) and its tangent at a point A is $2\mathrm{x}+3\mathrm{y}+9=0$

$\mathrm{P}(-1,2)\mathrm{and}\mathrm{Q}(-2,3)$
Hence image of point P from lie $2\mathrm{x}+3\mathrm{y}+9=0$
$\begin{array}{l}\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-2}{3}=-2\left(\frac{-2+6+9}{13}\right)\\ \mathrm{X}=-5; \mathrm{y}=-4\\ \mathrm{P}=(-5,-4)\end{array}$
Now length of PQ $=\sqrt{9+49}=\sqrt{58}=2\mathrm{a}\Rightarrow \mathrm{a}=\frac{\sqrt{58}}{2}$
We know that, $2\mathrm{ae}=\mathrm{PQ}$
$2\mathrm{ae}=\sqrt{2}\Rightarrow \mathrm{ae}=\frac{1}{\sqrt{2}}\Rightarrow {\mathrm{a}}^2{\mathrm{e}}^2=\frac{1}{2}$
${\mathrm{b}}^2={\mathrm{a}}^2(1-{\mathrm{e}}^2)$
${\mathrm{b}}^2=\frac{58}{4}-\frac{1}{2}\Rightarrow {\mathrm{b}}^2=\frac{58-2}{4}\Rightarrow {\mathrm{b}}^2=\frac{56}{4}\Rightarrow \mathrm{b}=\sqrt{14}$
${\mathrm{b}}^2=14$ length of minor axis $=2\sqrt{14}=\sqrt{56}$
Equation of line PQ is
$(\mathrm{y}-3)=\frac{4-3}{-3}(\mathrm{x}+2)$
$3\mathrm{y}–9=72+14$
$7\mathrm{x}–3\mathrm{y}+23=0$
On solving  we get $7\mathrm{x}–3\mathrm{y}+23=0\mathrm{and}2\mathrm{x}+3\mathrm{y}+9=0$ 
we get $\left(\frac{{\displaystyle -32}}{{\displaystyle 9}},\frac{{\displaystyle -17}}{{\displaystyle 27}}\right)$