A) \[{{d}^{2}}+{{(3b-2c)}^{2}}=0\]
B) \[{{d}^{2}}+{{(3b+2c)}^{2}}=0\]
C) \[{{d}^{2}}+{{(2b-3c)}^{2}}=0\]
D) \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\]
Correct Answer: D
Solution :
Two given parabola intersect at (0,0) and (4a, 4a). So, the common chord is y = x. On comparing y = x with given line, we get \[\frac{2b}{1}=\frac{3c}{-1}=\frac{4d}{0}\]\[\Rightarrow \]d = 0 and \[2b=3c=0\]You need to login to perform this action.
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