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 [2023]

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Solution

Q.
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 [2023]

1 $\frac{d}{a}, \frac{e}{b}, \frac{f}{c} are in G.P.$

2 $d, e, f are in A.P.$

3 $d, e, f are in G.P.$

4 $\frac{d}{a}, \frac{e}{b}, \frac{f}{c} are in A.P.$

Ans.
(4)

$a x^{2} + 2 b x + c y = 0$ ...(i)

$and d x^{2} + 2 e x + f y = 0$ ...(ii)

Equation (i) and (ii) intersect at $(α, 1)$ .

$∴$ $a α^{2} + 2 b α + c = 0$ ...(iii)

   $d α^{2} + 2 e α + f = 0$ ...(iv)

Roots of equation (iii):

$α = \frac{- 2 b \pm \sqrt{4 b^{2} - 4 a c}}{2 a} = - \frac{b}{a} [∵ a, b, c are in G.P. \Rightarrow b^{2} = a c]$

$α is also a root of equation (iv)$

$∴$    $d {(\frac{- b}{a})}^{2} + 2 e (\frac{- b}{a}) + f = 0$ $\Rightarrow \frac{d}{a} + \frac{f}{c} = \frac{2 e}{b}$

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

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