The points of contact of the tangents drawn from $\left(4,\sqrt{3}\right)$ to the ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{9}=1$ are

Given ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{9}=1$

Let $\mathrm{P}({\mathrm{x}}_1, {\mathrm{y}}_1)$ be point of contact 

Now equation of tangent at P is ${\mathrm{S}}_1=0$ 

$\frac{\mathrm{x} {\mathrm{x}}_1}{16}+\frac{\mathrm{y} {\mathrm{y}}_1}{9}=1$

If this tangent passes through $\left(4, \sqrt{3}\right)$ then

$\begin{array}{rcl}\frac{{\mathrm{x}}_1}{4}+\frac{{\mathrm{y}}_1}{3\sqrt{3}}& =& 1 \Rightarrow \frac{{\mathrm{y}}_1}{3\sqrt{3}}=1-\frac{{\mathrm{x}}_1}{4}\\ & \Rightarrow & {\mathrm{y}}_1=\frac{12\sqrt{3}-3\sqrt{3} {\mathrm{x}}_1}{4}\end{array}$

since P lies on the ellipse then $\frac{{{\mathrm{x}}_1}^2}{16}+\frac{{{\mathrm{y}}_1}^2}{9}=1$

$\begin{array}{rcl} & \Rightarrow & \frac{{{\mathrm{x}}_1}^2}{16}+\frac{{\left(12\sqrt{3}-3\sqrt{3} {\mathrm{x}}_1\right)}^2}{144}=1\\ & \Rightarrow & {{\mathrm{x}}_1}^2+{\left(4\sqrt{3}-\sqrt{3} {\mathrm{x}}_1\right)}^2=16\\ & \Rightarrow & {{\mathrm{x}}_1}^2+48+3{{\mathrm{x}}_1}^2-24{\mathrm{x}}_1=16\Rightarrow 4{{\mathrm{x}}_1}^2-24{\mathrm{x}}_1+32=0\Rightarrow {\mathrm{x}}_1=2 \left(\mathrm{or}\right) 4\\ & & \\ & & \end{array}$

If ${\mathrm{x}}_1=2 \Rightarrow {\mathrm{y}}_1=\frac{3\sqrt{3}}{2}$

If ${\mathrm{x}}_1=4 \Rightarrow {\mathrm{y}}_1=0$

$\therefore$ Points are $\left(2, \frac{3\sqrt{3}}{2}\right), (4, 0)$