The parabolas ax^2+2bx+cy=0 and dx^2+2ex+fy=0 intersect on the line y=1.

Question

The parabolas $ax^2+2bx+cy=0$ and $dx^2+2ex+fy=0$ intersect on the line $y=1$. If $a,b,c,d,e,f$ are positive real numbers and $a,b,c$ are in G.P., then:

Accepted Answer

At $y=1$: $ax^2+2bx+c=0$. Since $b^2=ac$ (G.P.), this factors as $(x\sqrt a+\sqrt c)^2=0$, giving the repeated root $x=-\sqrt{c/a}$.

Since this same $x$ satisfies $dx^2+2ex+f=0$: substituting $x=-\sqrt{c/a}$,
$$d\cdot\frac{c}{a} - 2e\sqrt{\frac{c}{a}} + f = 0 \implies \frac{d}{a}+\frac{f}{c} = \frac{2e}{\sqrt{ac}} = \frac{2e}{b}\ \ (\text{since } b=\sqrt{ac}).$$

So $\dfrac{d}{a}, \dfrac{e}{b}, \dfrac{f}{c}$ are in A.P.

Solutions · 1