The normals at three points P, Q, R of the pa | vedclass

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(A) The equation of any normal to the parabola $y^{2}=4ax$ is $y=mx-2am-am^{3}$.If it passes through $(h, k)$,then $k=mh-2am-am^{3}$,or $am^{3}+(2a-h)

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$y = 0$

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(A) The equation of any normal to the parabola $y^{2}=4ax$ is $y=mx-2am-am^{3}$.If it passes through $(h, k)$,then $k=mh-2am-am^{3}$,or $am^{3}+(2a-h)m+k=0$.This is a cubic equation in $m$. Let its roots be $m_{1}, m_{2}, m_{3}$.Then $m_{1}+m_{2}+m_{3}=0$ and $m_{1}m_{2}+m_{2}m_{3}+m_{3}m_{1}=\frac{2a-h}{a}$.The coordinates of $P, Q, R$,the feet of the three normals,are $(am_{1}^{2}, -2am_{1}), (am_{2}^{2}, -2am_{2}), (am_{3}^{2}, -2am_{3})$.Let $G(\bar{x}, \bar{y})$ be the centroid of $\Delta PQR$.Then $\bar{x} = \frac{a(m_{1}^{2}+m_{2}^{2}+m_{3}^{2})}{3} = \frac{a}{3}[(m_{1}+m_{2}+m_{3})^{2}-2(m_{1}m_{2}+m_{2}m_{3}+m_{3}m_{1})] = \frac{a}{3}[0 - 2(\frac{2a-h}{a})] = \frac{2}{3}(h-2a)$.And $\bar{y} = \frac{-2a(m_{1}+m_{2}+m_{3})}{3} = -\frac{2a}{3}(0) = 0$.Thus,the centroid of $\Delta PQR$ is $G(\frac{2}{3}(h-2a), 0)$,which lies on the line $y=0$.

The normals at three points $P, Q, R$ of the parabola $y^2 = 4ax$ meet in $(h, k).$ The centroid of triangle $PQR$ lies on :

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