Find the locus of the point of intersection o | vedclass

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(B) The given equation of the parabola is $y^2 - 6y + 24x - 63 = 0$.Completing the square for $y$,we get $(y - 3)^2 - 9 + 24x - 63 = 0$.$(y - 3)^2 = -

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$x - 9 = 0$

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(B) The given equation of the parabola is $y^2 - 6y + 24x - 63 = 0$.Completing the square for $y$,we get $(y - 3)^2 - 9 + 24x - 63 = 0$.$(y - 3)^2 = -24x + 72$.$(y - 3)^2 = -24(x - 3)$.This is a parabola of the form $(y - k)^2 = 4a(x - h)$,where $4a = -24$,so $a = -6$.The locus of the point of intersection of perpendicular tangents to a parabola is its directrix.The directrix of the parabola $(y - k)^2 = 4a(x - h)$ is given by $x - h = -a$.Here,$h = 3$,$k = 3$,and $a = -6$.So,$x - 3 = -(-6) = 6$.$x = 9$,which can be written as $x - 9 = 0$.

Find the locus of the point of intersection of perpendicular tangents to the parabola $y^2 - 6y + 24x - 63 = 0$.

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