JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 9)
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 :
$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.
$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in G.P.
$d, e, f$ are in A.P.
$d, e, f$ are in G.P.
Explanation
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.
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