Let the tangent and normal at any point P(at^ | vedclass

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(A) The coordinates of point $P$ are $(at^2, 2at)$.The equation of the tangent at $P$ is $ty = x + at^2$. Setting $y=0$, we get $x = -at^2$, so $T = (

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$a(1+t^2)$

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(A) The coordinates of point $P$ are $(at^2, 2at)$.The equation of the tangent at $P$ is $ty = x + at^2$. Setting $y=0$, we get $x = -at^2$, so $T = (-at^2, 0)$.The equation of the normal at $P$ is $y = -tx + 2at + at^3$. Setting $y=0$, we get $x = 2a + at^2$, so $G = (2a + at^2, 0)$.Since the tangent and normal are perpendicular, $\angle PTG = 90^\circ$, which implies that $TG$ is the diameter of the circle passing through $P, T,$ and $G$.The length of the diameter $TG = |(2a + at^2) - (-at^2)| = |2a + 2at^2| = 2a(1+t^2)$.Therefore, the radius of the circle is $\frac{1}{2} TG = a(1+t^2)$.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

Let the tangent and normal at any point $P(at^2, 2at)$, $(a > 0)$, on the parabola $y^2 = 4ax$ meet the axis of the parabola at $T$ and $G$ respectively. Then the radius of the circle through $P, T$ and $G$ is

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