Question

The asymptotes of the curve $x^2+4xy+3y^2+4x-3y+1=0$ passes through a fixed point $(h,k)$ then $h+k$ is

Answer 1

Let the equation of the asymptote be
$y=mx+c$
Putting this in the equation of the curve,
$x^2+4xy+3y^2+4x-3y+1=0\Rightarrow x^2+4x(mx+c)+3(mx+c{)}^2+4x-3(mx+c)+1=0\Rightarrow (1+4m+3m^2)x^2+(4c+6cm+4-3m)x+(3c^2-3c+1)=0\Rightarrow A{x}^2+Bx+C=0$

Asymptote meaning both roots at infinity,
Putting $x\to \frac{1}{t}$
$\Rightarrow C{t}^2+Bt+A=0$
This equation must have two zero roots,
$\therefore B=0\text{and}A=0;C\ne 0$
$C=(3c^2-3c+1)>0[\because D<0]$

Now, $A=0$
$\Rightarrow 1+4m+3m^2=0\Rightarrow (3m+1)(m+1)=0\Rightarrow m=-1,-\frac{1}{3}$
$B=04c+6cm+4-3m=0\Rightarrow c=\frac{3m-4}{6m+4}=\frac{7}{2},\frac{-5}{2}$

Equation of the asymptotes are,
$y=-x+\frac{7}{2}\cdots \left(1\right)y=-\frac{x}{3}-\frac{5}{2}\cdots \left(2\right)$
Finding point of intersection $(h,k)$,
$-h+\frac{7}{2}=-\frac{h}{3}-\frac{5}{2}\Rightarrow h=9,k=-\frac{11}{2}$

Hence, $h+k=3.5$