The line y = 6x + 1 touches the parabola y^2 | vedclass

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(C) The equation of the parabola is $y^2 = 24x$,which is of the form $y^2 = 4ax$,where $4a = 24$,so $a = 6$.The directrix of the parabola is $x = -a$,

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$(-6, -35)$

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(C) The equation of the parabola is $y^2 = 24x$,which is of the form $y^2 = 4ax$,where $4a = 24$,so $a = 6$.The directrix of the parabola is $x = -a$,which is $x = -6$.The locus of the point of intersection of two perpendicular tangents to a parabola is its directrix.We are looking for a point $P$ on the line $y = 6x + 1$ such that the tangent drawn from $P$ to the parabola is perpendicular to the given line.Since the point $P$ must lie on the directrix,we substitute $x = -6$ into the equation of the line $y = 6x + 1$.$y = 6(-6) + 1 = -36 + 1 = -35$.Thus,the coordinates of point $P$ are $(-6, -35)$.

The line $y = 6x + 1$ touches the parabola $y^2 = 24x$. The coordinates of a point $P$ on this line,from which the tangent to $y^2 = 24x$ is perpendicular to the line $y = 6x + 1$,is

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