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(D) The equation of a normal with slope $m$ to the parabola $y^2=4ax$ is given by $y=mx-2am-am^3$. If this normal passes through a point $(h, k)$,then

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$y^2-ax+3a^2=0$

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(D) The equation of a normal with slope $m$ to the parabola $y^2=4ax$ is given by $y=mx-2am-am^3$. If this normal passes through a point $(h, k)$,then $k=mh-2am-am^3$,which is a cubic equation in $m$: $am^3 + (2a-h)m + k = 0$. Let the roots of this equation be $m_1, m_2, m_3$. Since the normals are perpendicular,let $m_1m_2 = -1$. From the properties of roots,the product of roots is $m_1m_2m_3 = -k/a$. Substituting $m_1m_2 = -1$,we get $(-1)m_3 = -k/a$,so $m_3 = k/a$. Since $m_3$ is a root,it must satisfy the cubic equation: $a(k/a)^3 + (2a-h)(k/a) + k = 0$. Simplifying this: $k^3/a^2 + 2k - hk/a + k = 0$. Dividing by $k$ (assuming $k 
eq 0$): $k^2/a^2 + 3 - h/a = 0$. $k^2/a^2 = h/a - 3$. $k^2 = ah - 3a^2$. Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = ax - 3a^2$,or $y^2 - ax + 3a^2 = 0$.

The locus of the points of intersection of perpendicular normals to the parabola $y^2=4ax$ is

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