If $a \neq 0$ and the line 2bx + 3cy + 4d = 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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Solution

The correct option is A
The points of intersection are (0, 0), (4a, 4a)
(0, 0) lies on $2 b x + 3 c y + 4 d = 0 \Rightarrow d = 0$
(4a, 4a) lies on $2 b x + 3 c y + 4 d = 0 \Rightarrow 8 a b + 12 a c = 0 \Rightarrow 2 b + 3 c = 0 [∵ a \neq 0]$
$∴ d^{2} + (2 b + 3 c)^{2} = 0$

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If a≠0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2=4ax and x2=4ay , then

The correct option is A The points of intersection are (0, 0), (4a, 4a) (0, 0) lies on 2bx+3cy+4d=0⇒d=0 (4a, 4a) lies on 2bx+3cy+4d=0⇒8ab+12ac=0⇒2b+3c=0[∵a≠0] ∴d2+(2b+3c)2=0

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

The correct option is A
The points of intersection are (0, 0), (4a, 4a)
(0, 0) lies on $2 b x + 3 c y + 4 d = 0 \Rightarrow d = 0$
(4a, 4a) lies on $2 b x + 3 c y + 4 d = 0 \Rightarrow 8 a b + 12 a c = 0 \Rightarrow 2 b + 3 c = 0 [∵ a \neq 0]$
$∴ d^{2} + (2 b + 3 c)^{2} = 0$