If a0 and the line 2bx+3cy+4d=0 passes through the points of intersection of the parabolas…

Question

If $a
e0$ and the line $2bx+3cy+4d=0$ passes through the points of intersection of the parabolas $y^2=4ax$ and $x^2=4ay$, then:

Accepted Answer

Solving $y^2=4ax$ and $x^2=4ay$ simultaneously gives the intersection points $(0,0)$ and $(4a,4a)$.

Substituting $(0,0)$ into $2bx+3cy+4d=0$: $d=0$.

Substituting $(4a,4a)$: $2b(4a)+3c(4a)+4d=0\Rightarrow 8ab+12ac+4d=0$; with $d=0$ and $a
e0$: $2b+3c=0$.

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

Solutions · 1