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(A) Let the point be $(h, k)$. The equation of a normal to the parabola $y^{2} = 4ax$ is given by $y = mx - 2am - am^{3}$.Since it passes through $(h,

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$y^{2} = 4ax$

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(A) Let the point be $(h, k)$. The equation of a normal to the parabola $y^{2} = 4ax$ is given by $y = mx - 2am - am^{3}$.Since it passes through $(h, k)$,we have $k = mh - 2am - am^{3}$,which simplifies to $am^{3} + m(2a - h) + k = 0$.Let the slopes of the normals be $m_1, m_2, m_3$. The product of the roots is $m_1 m_2 m_3 = -k/a$.Given that the normals make angles $\alpha$ and $\beta$ with the axis,their slopes are $m_1 = 	an \alpha$ and $m_2 = 	an \beta$. We are given $m_1 m_2 = 2$.Substituting this into the product of roots,we get $2 m_3 = -k/a$,so $m_3 = -k/(2a)$.Since $m_3$ is a root of the cubic equation $am^{3} + m(2a - h) + k = 0$,we substitute $m = -k/(2a)$:$a(-k/(2a))^{3} + (-k/(2a))(2a - h) + k = 0$$-k^{3}/(8a^{2}) - k + kh/(2a) + k = 0$$-k^{3}/(8a^{2}) + kh/(2a) = 0$$k^{2} = 4ah$.Replacing $(h, k)$ with $(x, y)$,the locus is $y^{2} = 4ax$.

If two normals are drawn from a point to the parabola $y^{2} = 4ax$ such that they make angles $\alpha$ and $\beta$ with the axis,and $\tan \alpha \cdot \tan \beta = 2$,find the locus of the point.

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