The tangents drawn from the ends of the latus | vedclass

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(A) For the parabola ${y^2} = 4ax$,we have $4a = 12$,so $a = 3$.The coordinates of the ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are

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Directrix

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(A) For the parabola ${y^2} = 4ax$,we have $4a = 12$,so $a = 3$.The coordinates of the ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(3, 6)$ and $(3, -6)$.The equation of the tangent to the parabola ${y^2} = 4ax$ at point $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.For point $(3, 6)$,the tangent is $6y = 6(x + 3) \implies y = x + 3$.For point $(3, -6)$,the tangent is $-6y = 6(x + 3) \implies y = -(x + 3)$.Solving these two equations: $x + 3 = -(x + 3) \implies 2x + 6 = 0 \implies x = -3$.Since the equation of the directrix for ${y^2} = 12x$ is $x = -a$,which is $x = -3$,the tangents meet at the directrix.

The tangents drawn from the ends of the latus rectum of the parabola ${y^2} = 12x$ meet at:

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