Geometry or calculus? I

First part solving the two eqautions that is very easy.

$43-3{x}^{2}=128x-16{x}^{2}-224$ (Eliminating y from both the equations)

$\Rightarrow 13{x}^{2}-128x+267=0$

Solving $x=3,\frac{89}{13}$

But since $x \le \sqrt{\frac{43}{3}}$ hence only one solution is possible that is :

$x=3$

Solving for $y$ we get $y=2,-2$

Let's take the point $(3,2)$

Second part finding equation of tangents( that is finding their slopes)

We know that equation of a tangent to the conic $a{x}^{2}+2hxy+b{y}^{2}+2gx+2fy+c=0$ at a point $(u,v)$ is given by :

$T=0$ where :

$T=aux+h(uy+vx)+bvy+g(x+u)+f(y+v)+c$

Putting the co-ordinates and finding equation of tangent we get :

${T}_{3{x}^{2}+4{y}^{2}-43=0}=9x+8y-43=0$

${T}_{4{x}^{2}+{y}^{2}-32x+56=0} = 2x-y=4 = 0$

$\Rightarrow {m}_{1}=\frac{-9}{8}$

and ${m}_{2}=2$

Last part finding the angle between the tangents :

Let the acute angle between them be $\theta$ . Hence :

$tan(\theta)=\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right|$

Putting ${m}_{1}$ and ${m}_{2}$ we get :

$tan(\theta)=\frac{5}{2}$

Hence the obtuse angle is given by :

$=\pi { -tan }^{ -1 }(\frac { 5 }{ 2 } ) \approx 111.801$