One of the points on the parabola y^2 = 12x w | vedclass

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(B) The equation of the parabola is $y^2 = 12x$. Comparing this with $y^2 = 4ax$,we get $4a = 12$,so $a = 3$.For any point $P(x, y)$ on the parabola,t

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$(9, 6\sqrt{3})$

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(B) The equation of the parabola is $y^2 = 12x$. Comparing this with $y^2 = 4ax$,we get $4a = 12$,so $a = 3$.For any point $P(x, y)$ on the parabola,the focal distance is given by $x + a$.Given the focal distance is $12$,we have $x + 3 = 12$,which implies $x = 9$.Substituting $x = 9$ into the parabola equation $y^2 = 12x$,we get $y^2 = 12(9) = 108$.Thus,$y = \pm \sqrt{108} = \pm 6\sqrt{3}$.Therefore,the point is $(9, 6\sqrt{3})$ or $(9, -6\sqrt{3})$.Comparing this with the given options,the correct point is $(9, 6\sqrt{3})$.

One of the points on the parabola $y^2 = 12x$ with focal distance $12$ is:

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