The parabolas : a x 2 + 2 b x + c y = 0 and d 2 + 2 e x + f y = 0 intersect on the line y = 1 . If a ,  …
The parabolas : and intersect on the line . If are positive real numbers and are in , then
are in A.P.
are in G.P.
are in A.P.
are in G.P.
Solution
Given, $ax^2 + 2bx + cy = 0$ and $d^2 + 2ex + fy = 0$ intersect on the line $y = 1$,
And $a$, $b$, $c$ are in G.P.
So, $b^2 = ac$
Now putting the value of $b$ in $ax^2 + 2bx + cy = 0$ and taking $y = 1$ we get,
$ax^2 + 2bx + c = 0$
$\Rightarrow ax^2 + 2\sqrt{ac}x + c = 0 \Rightarrow b^2 = ac$
$\Rightarrow (x\sqrt{a} + \sqrt{c})^2 = 0$
$\Rightarrow x = -\frac{\sqrt{c}}{\sqrt{a}}$ and $x^2 = $\frac{c}{a}$...1$
Now, putting the value of $x$ and $x^2$ in $d^2 + 2ex + fy = 0$ and taking $y = 1$ we get,
$\Rightarrow dc + 2e\left(-\frac{\sqrt{c}}{\sqrt{a}}\right) + f = 0$
$\Rightarrow\frac{dc}{a} + f = 2e\sqrt{\frac{c}{a}}$
$\Rightarrow \frac{d}{a} + \frac{f}{c} = 2e\sqrt{\frac{1}{ac}}$
$\Rightarrow$ $\frac{d}{a}$ + $\frac{f}{c}$ = $\frac{2e}{b}$ as $b$ = $\sqrt{ac}$
$\therefore \frac{d}{a}$, $\frac{e}{b}$, $\frac{f}{c}$ are in A.P.