"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 :"

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Accepted Answer

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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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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 :

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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 :

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