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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 \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f} are positive real numbers and a, b, c are in G.P., then

Option: 1

d, e, f are in G.P.


Option: 2

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


Option: 3

d, e, f are in A.P.


Option: 4

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


Answers (1)

best_answer

at y=1, Both curve intersect

\left.\Rightarrow \begin{array}{l} a x^{2}+2 b x+c=0 \\ d x^{2}+2 \mathrm{ex}+\mathrm{f}=0 \end{array}\right\} \text { Common Root }

Given a, b, c are in G.P
b^{2}=a c

\Rightarrow D=4 b^{2}-4 a c=0 for the first equation
\Rightarrow Both the Root are equal
 \because \text{sum of the roots}=-2 \frac{b}{a}
\alpha+\alpha=-2 \frac{b}{a}
\alpha=-\frac{b}{a}
It satisfies the second equation also

d\left(-\frac{b}{a}\right)^{2}+2 e\left(-\frac{b}{a}\right)+f=0
d\left(\frac{b^{2}}{a^{2}}\right)-\frac{2 e b}{a}+f=0
d\left(\frac{a c}{a^{2}}\right)-2 e \frac{b}{a}+f=0

\frac{d}{a}-\frac{2 e b}{a c}+\frac{f}{c}=0
\frac{d}{a}-\frac{2 e b}{b^{2}}+\frac{f}{c}=0

\Rightarrow 2 \frac{e}{b}=\frac{d}{a}+\frac{f}{c} \Rightarrow \frac{d}{a}, \frac{e}{b}, \frac{f}{c} \text { are\, in AP }
 

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