meet at (h, k). The centroid of the 
lies on
A) x = 0
B) y = 0
C) x = -a
D) y = a
Correct Answer: B
Solution :
We know that, the sum of ordinates of feet of normals drawn from a point to the parabola, \[{{E}_{2}}={{V}_{2}}{{l}_{2}}{{t}_{2}}\] is always zero. Now, as normals at three points P, Q AND R of parabola \[=14\times 5\times 15\] meet at (h , k). \[\eta =\frac{{{E}_{2}}}{{{E}_{1}}}\times 100\]The normals from (h, k) to \[=0.875\times 100=87.5%\]meet the parabola at P, Q and R. \[\frac{{{B}_{centre}}}{{{B}_{axis}}}={{\left( 1+\frac{{{X}^{2}}}{{{R}^{2}}} \right)}^{3/2}}\] y ? coordinates \[{{B}_{axis}}=\frac{1}{8}{{B}_{centre}}\] of these points P, Q and R will be zero. \[\frac{8}{1}={{\left( 1+\frac{{{X}^{2}}}{{{R}^{2}}} \right)}^{3/2}}\] y ? coordinates of the centroid of \[4=1+\frac{{{X}^{2}}}{{{R}^{2}}}\] \[3=\frac{{{X}^{2}}}{{{R}^{2}}}\] Hence, centroid lies on y = 0.
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