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Question

If $$a\neq 0$$ and the line $$2bx+3cy+4d=0$$ passes through the points of intersection of the parabolas $$y^{2}=4ax$$ and $$x^{2}=4ay,$$ then


A
d2+(2b+3c)2=0
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B
d2+(3b+2c)2=0
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C
d2+(2b3c)2=0
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D
d2+(3b2c)2=0
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Solution

The correct option is A $$d^{2}+(2b+3c)^{2}=0$$
The point of intersection of $$y^2=4ax$$ and $$x^2=4ay$$ is $$(0,0)$$ and $$(4a,4a)$$
Therefore, given line is same as $$y=x$$
Slope$$=\dfrac { -2b }{ 3c } =1$$
$$\Rightarrow 2b+3c=0$$      ...(1)
Also line passes through $$(0,0)$$
Therefore, $$d=0$$       ...(2)
From (1) and (2): $$d^{2}+(2b+3c)^{2}=0$$
Ans: A

Mathematics

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