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 :
Given prarbolas are \[{{y}^{2}}=4ax\] .....(i) \[{{x}^{2}}=4ay\] .....(ii) Putting the value of y from (ii) in (i), we get \[\frac{{{x}^{4}}}{16{{a}^{2}}}=4ax\Rightarrow x({{x}^{3}}-64{{a}^{3}})=0\Rightarrow x=0,\,4a\]. from (ii), \[y=0,\,4a\]. Let \[A\equiv (0,\,0);\,B\equiv (4a,\,4a)\] Since, given line \[2bx+3cy+4d=0\] passes through A and B, \\[d=0\] and \[8ab+12ac=0\Rightarrow 2b+3c=0\],(\[\because \] \[a\ne 0\]) Obviously, \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\].
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