The locus of the point of intersection of any two perpendicular tangents to the parabola $x^{2} - 6 x + 16 y + 41 = 0$ is

4 y + 9 = 0

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y - 2 = 0

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y - 3 = 0

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6 y - 13 = 0

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Solution

The correct option is B $y - 2 = 0$
Here given parabola can be written as $(x - 3)^{2} = - 16 (y + 2)$
For a parabola $x^{2} = - 4 a y$ , the locus of points of intersection of perpendicular tangents is its directrix which is $y = a$
Thus, the given locus is
$y + 2 = 4 ∴ y = 2$

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Perpendicular Distance of a Point from a Line

Standard XII Mathematics

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The locus of the point of intersection of any two perpendicular tangents to the parabola x2−6x+16y+41=0 is

The correct option is B y−2=0Here given parabola can be written as (x−3)2=−16(y+2) For a parabola x2=−4ay, the locus of points of intersection of perpendicular tangents is its directrix which is y=a Thus, the given locus is y+2=4∴y=2

The locus of the point of intersection of any two perpendicular tangents to the parabola $x^{2} - 6 x + 16 y + 41 = 0$ is

The correct option is B $y - 2 = 0$
Here given parabola can be written as $(x - 3)^{2} = - 16 (y + 2)$
For a parabola $x^{2} = - 4 a y$ , the locus of points of intersection of perpendicular tangents is its directrix which is $y = a$
Thus, the given locus is
$y + 2 = 4 ∴ y = 2$