If the normals of the parabola y^2 = 4x drawn | vedclass

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(B) The parabola is $y^2 = 4x$,so $a = 1$. The end points of the latus rectum are $(1, 2)$ and $(1, -2)$.The equation of the normal to the parabola $y

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$2$

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(B) The parabola is $y^2 = 4x$,so $a = 1$. The end points of the latus rectum are $(1, 2)$ and $(1, -2)$.The equation of the normal to the parabola $y^2 = 4ax$ at $(at^2, 2at)$ is $y = -tx + 2at + at^3$. For $y^2 = 4x$,$a = 1$,so the normal is $y = -tx + 2t + t^3$.At $(1, 2)$,$t = 1$,so the normal is $y = -x + 2 + 1$,which simplifies to $x + y = 3$.At $(1, -2)$,$t = -1$,so the normal is $y = -(-1)x + 2(-1) + (-1)^3$,which simplifies to $y = x - 3$,or $x - y = 3$.The normals are $x + y - 3 = 0$ and $x - y - 3 = 0$. These lines are tangents to the circle $(x - 3)^2 + (y + 2)^2 = r^2$ with center $(3, -2)$.The perpendicular distance from the center $(3, -2)$ to the line $x + y - 3 = 0$ is $r = \frac{|3 + (-2) - 3|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.Thus,$r^2 = 2$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

If the normals of the parabola $y^2 = 4x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^2 + (y + 2)^2 = r^2$,then the value of $r^2$ is

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