The equation of a tangent to the parabola y^2 | vedclass

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(D) The equation of the parabola is $y^2 = 8x$,which is of the form $y^2 = 4ax$,where $4a = 8$,so $a = 2$.The directrix of the parabola $y^2 = 4ax$ is

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$(-2, 0)$

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(D) The equation of the parabola is $y^2 = 8x$,which is of the form $y^2 = 4ax$,where $4a = 8$,so $a = 2$.The directrix of the parabola $y^2 = 4ax$ is given by $x = -a$,which is $x = -2$.It is a standard property of parabolas that the locus of the intersection of two perpendicular tangents is the directrix of the parabola.Since the point lies on the given tangent line $y = x + 2$ and also on the directrix $x = -2$,we substitute $x = -2$ into the line equation:$y = (-2) + 2 = 0$.Thus,the required point is $(-2, 0)$.

The equation of a tangent to the parabola $y^2 = 8x$ is $y = x + 2$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

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