P is a point on the parabola y^2 = 4ax (a 0 | vedclass

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(D) Let the point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The vertex $A$ is $(0, 0)$.The equation of the line $PA$ passing through $(0, 0)$ a

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Both $(A)$ and $(B)$

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(D) Let the point $P$ on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The vertex $A$ is $(0, 0)$.The equation of the line $PA$ passing through $(0, 0)$ and $(at^2, 2at)$ is $y = \frac{2}{t}x$.The directrix of the parabola is $x = -a$.To find $D$,substitute $x = -a$ into the line equation: $y = \frac{2}{t}(-a) = -\frac{2a}{t}$. So,$D = (-a, -\frac{2a}{t})$.$M$ is the foot of the perpendicular from $P(at^2, 2at)$ to the directrix $x = -a$,so $M = (-a, 2at)$.The circle with $MD$ as diameter has the equation $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Here,$(x_1, y_1) = (-a, 2at)$ and $(x_2, y_2) = (-a, -\frac{2a}{t})$.$(x + a)(x + a) + (y - 2at)(y + \frac{2a}{t}) = 0$.To find the intersection with the $x$-axis,set $y = 0$:$(x + a)^2 + (-2at)(\frac{2a}{t}) = 0$.$(x + a)^2 - 4a^2 = 0$.$(x + a)^2 = 4a^2$.$x + a = \pm 2a$.$x = a$ or $x = -3a$.Thus,the coordinates are $(a, 0)$ and $(-3a, 0)$.

$P$ is a point on the parabola $y^2 = 4ax$ $(a > 0)$ whose vertex is $A$. $PA$ is produced to meet the directrix in $D$ and $M$ is the foot of the perpendicular from $P$ on the directrix. If a circle is described on $MD$ as a diameter,then it intersects the $x$-axis at a point whose coordinates are:

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