If $a\not =0$ and the line $2bx+3cy+4d=0$ passes through the points of intersection of the parabolas $y^2=4ax$ and $x^2=4ay$

Question

If $a\not =0$ and the line $2bx+3cy+4d=0$ passes through the points of intersection of the parabolas $y^2=4ax$ and $x^2=4ay$, find relation between $b,c$ and $d$.

Since both the parabolas are symmetric, $x=y$

I found the point of intersections to be $(0,0)$ and $(4a,4a)$

So in the equation of line $d=0$

And $$2bx+3cy=0$$ $$8abx+12acy=0$$

Now this is where I got confused. What I did was put $x=y$

So I got the relation $2b=-3c$

Now the question had options in it, so I tested each option out to obtain the right answer

$$d^2+(2b+3c)^2=0$$

But, if I was in a situation where the opinions were not so easy to interpret, or if there were none at all, how should I solve it?

Answer 1

The objective of the problem is to express the relationship among b, c and d in a single equation. Note that you can always combine $n$ equations $f_i = 0$ for $i = 1,2,...n$ into one equation by setting their quadratic sum to zero, i.e.

$$f_1^2 + f_2^2 + … + f_n^2 = 0$$

In your case, you have two equations,

$$f_1=d=0,\>\>\>\>\>f_2=2b+3c=0$$

which can then be written equivalently as

$$d^2+(2b+3c)^2=0$$