The parabolas : a x^2+2 b x+c y=0 and d x^2+2 e x+f y=0 intersect on the line y=1.

Question

The parabolas : $a x^2+2 b x+c y=0$ and $d x^2+2 e x+f y=0$ 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

Let point of intersection be ($$\alpha,1$$)

$$\alpha x^2+2b\alpha+c=0$$ ..... (i)

and $$d\alpha^2+2e\alpha+f=0$$ .... (ii)

$$\Rightarrow a\alpha^2+2\sqrt{ac}\alpha+c=0$$ ($$\because$$ $b^2=ac$)

$${\left( {\sqrt a \alpha  + \sqrt c } \right)^2} = 0$$

$$\alpha  =  - \sqrt {{c \over a}} $$

Put the value of $$\alpha$$ in (ii),

$$d{c \over a} - 2e\sqrt {{c \over a}}  + f = 0$$

$${d \over a} - {{2e} \over {\sqrt {ac} }} + {f \over c} = 0$$

$${d \over a} + {f \over c} = 2{e \over b}$$

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

Solutions · 1