$y^2=4ax$ will be a parabola and will intersect the given line at two points.
To be honest, that’s all I could figure out. I tried using $x=y^2/4a$ and substituting in the given equation, but the process was impossibly long.
What is the proper way to solve it?
On
Hint:
The equation of any quadratic curve passing through the intersection of $$lx+my+n=0$$ and $$y^2=4ax$$
can be chosen as $$y^2-4ax+K(lx+my+n)^2=0$$ where $K$ is an arbitrary constant which(may not be real ) can be determined using
What is condition for second degree equation to represent a pair of straight lines?
You may simplify the process by solving for $\frac yx$ directly since $\frac yx = k $ represents lines passing the origin. To do so, rewrite the two equation as
$$\frac yx = \frac{4a}y,\>\>\>\>\>\>\>m\frac yx +l=-\frac nx $$
Take their ratio to get the quadratic equation in $\frac yx$,
$$ \frac{y^2}{x^2}+\frac{4am}n\frac yx+ \frac{4al}n=0$$
Then, solve to obtain the pair of lines,
$$\frac yx = -\frac{2am}n \pm 2\sqrt{\frac{a^2m^2}{n^2}-\frac{al}n}$$
(Assuming $n\ne 0$ and the condition $\frac{a^2m^2}{n^2}>\frac{al}n$ holds for the two curve having intersections.)