A point P moves such that the sum of the angl | vedclass

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(A) Let the parabola be $y^2 = 4ax$. The equation of a normal to the parabola is $y = mx - 2am - am^3$. If the normal passes through $P(h, k)$,then $k

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a straight line

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(A) Let the parabola be $y^2 = 4ax$. The equation of a normal to the parabola is $y = mx - 2am - am^3$. If the normal passes through $P(h, k)$,then $k = mh - 2am - am^3$,which simplifies to $am^3 + (2a - h)m + k = 0$. Let the roots of this cubic equation be $m_1, m_2, m_3$. These represent the slopes of the three normals. The angles these normals make with the axis of the parabola (the $x$-axis) are $	heta_1, 	heta_2, 	heta_3$,where $	an 	heta_i = m_i$. The sum of the angles is $	heta_1 + 	heta_2 + 	heta_3 = C$ (constant). Taking the tangent on both sides,$	an(	heta_1 + 	heta_2 + 	heta_3) = 	an C$. Using the identity $	an(	heta_1 + 	heta_2 + 	heta_3) = \frac{S_1 - S_3}{1 - S_2}$,where $S_1 = \sum m_i = 0$,$S_2 = \sum m_i m_j = \frac{2a - h}{a}$,and $S_3 = m_1 m_2 m_3 = -\frac{k}{a}$. Substituting these,$\frac{0 - (-k/a)}{1 - (2a - h)/a} = 	an C$. This simplifies to $\frac{k/a}{(a - 2a + h)/a} = 	an C$,so $\frac{k}{h - a} = 	an C$. Thus,$k = (h - a) 	an C$,which represents a straight line passing through $(a, 0)$.

$A$ point $P$ moves such that the sum of the angles which the three normals drawn from $P$ to the standard parabola $y^2 = 4ax$ make with the axis of the parabola is constant. Then the locus of $P$ is:

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