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$y^2=4ax$ 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?

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  • $\begingroup$ What guarantee do you have that there are two intersection points? Without further conditions the line might be tangent to the parabola or not intersect it at all. $\endgroup$ – amd Jan 7 '20 at 17:54
  • $\begingroup$ I don't, but it seems obvious $\endgroup$ – Aditya Jan 7 '20 at 17:57
  • $\begingroup$ doubtnut.com/question-answer/… $\endgroup$ – lab bhattacharjee Jan 7 '20 at 18:00
  • $\begingroup$ Is it? It’s not to me, but perhaps there’s some other context that you haven’t included here that allows you to assume that not only are there two intersection points, but they generate two distinct lines (not the case when $n=0$). If I were grading this, I’d take marks off if you didn’t at least state the conditions under which your solution was valid. $\endgroup$ – amd Jan 7 '20 at 18:42
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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.)

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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?

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  • $\begingroup$ Your last equation is still that of a parabola. You need to use the double line $(lx+my+n)^2=0$ to change the conic’s type. If we allow $n=0$, then I believe that you need an linear (or affine) combination of the two conics as well, since in that case that double line is the unique line that passes through the two intersection points. $\endgroup$ – amd Jan 7 '20 at 17:53
  • $\begingroup$ @amd, Thanks for your observation $\endgroup$ – lab bhattacharjee Jan 7 '20 at 17:57

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