The point on the parabola y^2 = 8x at which t | vedclass

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(A) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The slope of the normal to the parabola $y^2 = 4ax$ at point $(x_1, y_1

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$(6, -4\sqrt{3})$

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(A) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The slope of the normal to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is given by $m = -\frac{y_1}{2a}$.The normal is inclined at $60^\circ$ to the $x$-axis,so its slope $m = 	an(60^\circ) = \sqrt{3}$.Equating the slopes: $\sqrt{3} = -\frac{y_1}{2(2)} = -\frac{y_1}{4}$.Thus,$y_1 = -4\sqrt{3}$.Since the point lies on the parabola $y^2 = 8x$,we have $(-4\sqrt{3})^2 = 8x_1$.$16 	imes 3 = 8x_1 \implies 48 = 8x_1 \implies x_1 = 6$.Therefore,the required point is $(6, -4\sqrt{3})$.

The point on the parabola $y^2 = 8x$ at which the normal is inclined at $60^\circ$ to the $x$-axis has the coordinates:

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