Fun With Calculus 1

$y = \dfrac{1+3x^{2}}{3+x^{2}} = 3 - \dfrac{8}{3+x^{2}}$
Substituting y = 1, we get $x = \pm 1$
Differentiating,
$\dfrac{dy}{dx} = \dfrac{16x}{(3+x^{2})^{2}}$
$y-1=-1(x+1) \to y = -x$
The point $(a,b)$ lines on $y = -x$
$\therefore a+b = 0$
This can be confirmed as the equation of other tangent is $y = x$ .
And the intersection is $(0,0)$

It is an even function thus for y=q solution is both 1, -1. It can be proved that tangents at these any two such point on an even function intersect.

$y=\dfrac{1+3x^2}{3+x^2}~\implies~if~y=1,~3+x^2=1+3x^2.~~So~2=2x^2.\\ \therefore~x = \pm 1.~~~and~~tangents~are~from~(1,1)~and~(-1,1).\\ Slope~of~tangents~to~y=\dfrac{1+3x^2}{3+x^2},~~is\\ y '=\dfrac{6x(3+x^2) - (1+3x^2)*2x}{(1+3x^2)^2}=\dfrac{16x}{9x^4+6x^2+1}.\\ \therefore~m_{x=1}=\dfrac{16}{9+6+1}=1.~~and~~m_{x=-1}=\dfrac{ - 16}{9+6+1}= - 1.\\ So~the~two~tangents~are~~~~y=1*(x-1)+1=x,~~~and~~y=-1(x+1)+1= - x.\\ These~are~two~lines~FROM~(0,0)~at~45^o~~and~~135^0. That~is~(a,b)=(0,0).\\ a+b=0+0=0.$