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anchit saini (4396)

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\mbox{Point of intersection of normals (y) =}-at_1t_2(t_1 + t_2) \\ \\<br/>=-a \ \mbox{given as it lies on y=-a} \\ \\<br/>\mbox{Hence , }t_1t_2(t_1 + t_2)=1 \\ \\<br/>\mbox{Also,pt of intersection of tangents is }-> \\ \\<br/>x= at_1t_2 \\ \\<br/>y= a(t_1 + t_2) \\ \\<br/>Thus, -><br/>xy=a^2t_1t_2(t_1 + t_2)=a^2


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