A) \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\]
B) \[{{d}^{2}}+{{(3b+2c)}^{2}}=0\]
C) \[{{d}^{2}}+{{(2b-3c)}^{2}}=0\]
D) \[{{d}^{2}}+{{(3b-2c)}^{2}}=0\]
Correct Answer: A
Solution :
The equation of parabolas are \[{{y}^{2}}=4ax\] and\[{{x}^{2}}=4ay\] On solving above curves, we get \[x=0\]and\[x=4a\] Also, \[y=0\]and\[y=4a\] \[\therefore \]The point of intersection of parabolas are \[A(0,0)\]and \[B(4a,\,4a)\]. Also, line\[2bx+3cy+4d=0\]passes through a and B, respectively. \[\therefore \] \[d=0\] ...(i) and \[2b.4a+3c.4a+4d=0\] \[\Rightarrow \] \[2ab+3ac+d=0\] \[\Rightarrow \] \[a(2b+3c)=0\] \[(\because \,d=0)\] \[\Rightarrow \] \[2b+3c=0\] ...(ii) On squaring and adding Eqs. (i) and (ii), we get \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\]
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