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(A) The equation of the parabola is $y^2 = 4ax$. The vertex is at $(0, 0)$.$A$ circle of radius $a$ touching the parabola at the vertex $(0, 0)$ must

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$\sqrt{3}a$

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(A) The equation of the parabola is $y^2 = 4ax$. The vertex is at $(0, 0)$.$A$ circle of radius $a$ touching the parabola at the vertex $(0, 0)$ must have its center at $(a, 0)$ or $(-a, 0)$. Given the parabola opens to the right,the circle is $(x - a)^2 + y^2 = a^2$.The endpoints of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.Let the point be $P = (a, 2a)$.The length of the tangent from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $\sqrt{S(x_1, y_1)}$.For the circle $(x - a)^2 + y^2 - a^2 = 0$,substituting $P(a, 2a)$:$L = \sqrt{(a - a)^2 + (2a)^2 - a^2} = \sqrt{0 + 4a^2 - a^2} = \sqrt{3a^2} = \sqrt{3}a$.

Find the length of the tangent drawn from an endpoint of the latus rectum of the parabola $y^2 = 4ax$ to a circle of radius $a$ that touches the parabola at its vertex.

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