Find the asymptotes to the hyperbola $3x^2+2xy-y^2+8x+10y+14=0$

Here is a view I have on hyperbola's asymptotes :

The asymptotes and the hyperbola behave the same way as we go to a point $(x,y)$ at an infinite distance from the centre of hyperbola, along the asymptotes.

So for a hyperbola: $H:ax^2+by^2+2hxy+2gx+2fy+c=0$, the terms containing $x$ and $y$ will affect the behaviour at such a point. Since the two(equation of hyperbola and the pair of asymptotes) have same behaviour (and hence same value of power of point), the only thing that differs in the two equations is the constant term.

So the equation of pair of asymptotes will be of the form: $A=ax^2+by^2+2hxy+2gx+2fy+d=0$

Now you can find the equation of asymptotes by using the fact that the centre of hyperbola passes lies on the asymptotes.

The centre can easily be found by solving: $\frac{\delta H}{\delta x}=0$ and $\frac{\delta H}{\delta y}=0$ .

So this point (Say $(x_o,y_o)$) will satisfy $A=0$

Find $d$ by putting the point in the equation.

For eg. in the hyperbola in your question:

The centre is $(-9/4,11/4)$ .

So if the equation of asymptotes is : $3x^2+2xy-y^2+8x+10y+\lambda=0$

To find $\lambda$ put the centre into the equation

$$\frac{243}{16}-\frac{99}{8} -\frac{121}{16}-18+\frac{55}{2}+\lambda=0$$

$$\lambda=\frac{19}{4}$$

So factorise the equation for asymptotes to get asymptotes.

$$(6x-2y+19)(2x+2y-1)=0$$

To justify the approach, note that asymptotically the quadratic terms of the curve $f(x,y)=3x^2+2xy-y^2+8x+10y+14$ dominate, i.e.

$$3x^2+2xy-y^2=(3x-y)(x+y)=0$$

which corresponds to the asymptotic behaviors of the asymptotes and yields their slopes $3,\> -1$.

To obtain the actual equations of the two asymptotes, let $f’_x= f’_y=0$ to determine the center, i.e.

$$ 6x+2y+8=0,\>\>\>\>\>2x-2y+10=0$$

Solve to get the center $(-\frac94, \frac{11}4)$. Then, use the point-slope formula for the equations

$$y-\frac{11}4=-(x+\frac94),\>\>\>\>\> y-\frac{11}4=3(x+\frac94)$$