3x² + 4y² = 12

$\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{3}=1$

$\mathrm{x}=2\mathrm{cos\theta },\mathrm{y}=\sqrt{3}\mathrm{sin\theta }$

$\mathrm{Equation}\mathrm{of}\mathrm{normal}\mathrm{is}\frac{{\mathrm{a}}^2\mathrm{x}}{{\mathrm{x}}_1}-\frac{{\mathrm{b}}^2\mathrm{y}}{{\mathrm{y}}_1}={\mathrm{a}}^2-{\mathrm{b}}^2$

$2\mathrm{xsin\theta }-\sqrt{3}\mathrm{cos\theta y}=\mathrm{sin\theta cos\theta }$

$\mathrm{Slope}\frac{2}{\sqrt{3}}\mathrm{tan\theta }=-2 \therefore \mathrm{tan\theta }=-\sqrt{3}$

Equation of tangent is it passes through (4, 4)

$3\mathrm{xcos\theta }+2\sqrt{3}\mathrm{sin\theta y}=6$

$12\mathrm{cos\theta }+8\sqrt{3}\mathrm{sin\theta }=6$

$\mathrm{cos\theta }=-\frac{1}{2},\mathrm{sin\theta }=\frac{\sqrt{3}}{2}\therefore \mathrm{\theta }=120\mathrm{°}$

$\mathrm{P}(-1,\frac{2}{2}),\mathrm{Q}(4,4)$

$\mathrm{PQ}=\frac{5\sqrt{5}}{2}$