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If $a \neq 0$ and the line $2 b x+3 c y+4 d=0$, passes through the points of intersection of the parabolas $y^2=4 a x$ and $x^2=4 a y$, then
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The correct answer is:
$d^2+(2 b+3 c)^2=0$
$y^2=4 a x, x^2=4 a y$ clearly they are symmetric
w.r.t $y=x$.
$\therefore y=x$ also pass through their point of intersections.
i.e., solving $y^2=4 a x$ and $y=x$
$\Rightarrow x^2=4 a x \Rightarrow x=0,4 a$
When $x=0, y=0 ; x=4 a \Rightarrow y=4 a$
$\therefore(0,0) \&(4 a, 4 a)$ are the points of intersection.
$\because 2 b x+3 c y+4 d=0$ passes through $(0,0) \&(4 a, 4 a)$ (given)
$\Rightarrow d=0$ and $(2 b+3 c)=0$
$\therefore d^2+(2 b+3 c)^2=0$
w.r.t $y=x$.
$\therefore y=x$ also pass through their point of intersections.
i.e., solving $y^2=4 a x$ and $y=x$
$\Rightarrow x^2=4 a x \Rightarrow x=0,4 a$
When $x=0, y=0 ; x=4 a \Rightarrow y=4 a$
$\therefore(0,0) \&(4 a, 4 a)$ are the points of intersection.
$\because 2 b x+3 c y+4 d=0$ passes through $(0,0) \&(4 a, 4 a)$ (given)
$\Rightarrow d=0$ and $(2 b+3 c)=0$
$\therefore d^2+(2 b+3 c)^2=0$
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