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 these, 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. \[\therefore \] \[d=0\] ...(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 Eqs. (i) and (ii) and then adding, we get \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\]You need to login to perform this action.
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