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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

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                    Equation of the line passing though the points of intersection of parabola \[{{x}^{2}}=8y\] and the ellipse\[\frac{{{x}^{2}}}{3}+{{y}^{2}}=1\] is:                   JEE Main Online Paper (Held On 09 April 2013)

    A)                 \[y-3=0\]                

    B)                 \[y+3=0\]                

    C)                 \[3y+1=0\]                

    D)                 \[3y-1=0\]                

    Correct Answer: D

    Solution :

                    On solving both the equations, we get                 \[\frac{8y}{3}+{{y}^{2}}=1\] \[\Rightarrow \]               \[3{{y}^{2}}+8y-3=0\] \[\Rightarrow \]               \[(3y-1)\,(y+3)=0\] \[\Rightarrow \]               \[y=-3,\,\,\frac{1}{3},\] here \[y\ne -3\] \[\therefore \]  At \[y=\frac{1}{3}\] \[\Rightarrow \]               \[x=\pm \,2\sqrt{\frac{2}{3}}\] \[\therefore \] Point of intersection is \[\left( 2\sqrt{\frac{2}{3}},\,\frac{1}{3} \right)\] and \[\left( -2\sqrt{\frac{2}{3}},\,\frac{1}{3} \right)\]                 From option (d); \[3y-1=0\] is the required equation which satisfied the intersection points.                


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