Locus of the point of intersection of the perpendiculars tangent of the curve $y^{2} + 4 y - 6 x - 2 = 0$ is

2 x - 1 = 0

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2 x + 3 = 0

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2 y + 3 = 0

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2 x + 5 = 0

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Solution

The correct option is D $2 x + 5 = 0$
Given parabola is, $y^{2} + 4 y - 6 x - 2 = 0$

$\Rightarrow y^{2} + 4 y + 4 = 6 x + 6 = 6 (x + 1)$

$\Rightarrow (y + 2)^{2} = 6 (x + 1)$

shifting origin to $(- 1, - 2)$

$Y^{2} = 4 a X$ where $a = \frac{3}{2}$

We know locus of point of intersection of perpendicular tangent is directrix of the parabola itself

Hence required locus is $X = - a \Rightarrow x + 1 = - \frac{3}{2} \Rightarrow 2 x + 5 = 0$

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Locus of the point of intersection of the perpendiculars tangent of the curve y2+4y−6x−2=0 is

Locus of the point of intersection of the perpendiculars tangent of the curve $y^{2} + 4 y - 6 x - 2 = 0$ is