The locus of the point of intersection of the | vedclass

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(D) The given equation is $y^2 + 4y - 6x - 2 = 0$.Completing the square for $y$,we get $(y^2 + 4y + 4) - 4 - 6x - 2 = 0$,which simplifies to $(y + 2)^

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$2x + 5 = 0$

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(D) The given equation is $y^2 + 4y - 6x - 2 = 0$.Completing the square for $y$,we get $(y^2 + 4y + 4) - 4 - 6x - 2 = 0$,which simplifies to $(y + 2)^2 = 6x + 6 = 6(x + 1)$.Let $Y = y + 2$ and $X = x + 1$. The equation becomes $Y^2 = 6X$.This is a parabola of the form $Y^2 = 4aX$,where $4a = 6$,so $a = 1.5$.The locus of the point of intersection of perpendicular tangents to a parabola is its directrix.The directrix of the parabola $Y^2 = 4aX$ is $X = -a$.Substituting back,we get $x + 1 = -1.5$,which implies $x + 1 = -3/2$.Multiplying by $2$,we get $2x + 2 = -3$,or $2x + 5 = 0$.

The locus of the point of intersection of the perpendicular tangents to the curve $y^2 + 4y - 6x - 2 = 0$ is:

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