• # question_answer If $a\ne 0$ and the line $2bx+3cy+4d=0$ passes through the points of intersection of the parabolas ${{y}^{2}}=4ax$ and ${{x}^{2}}=4ay$, then [AIEEE 2004] 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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