The equation of tangent to the parabola y 2-6 x at point 6-6 is A- 2 x-3 y-30-0B- 2 x - y -6-0C- 2 y - x -6-0D- x-y-12-0

The correct option is C 2y x 6=0Given parabola, y^2=6x The equation of tangent at (6,6) is nbsp; T = 0 ⇒ y·6=6(x+62) ⇒ 12y=6x+36 Hence, the required equatio ...

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Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

The equation ...

Question

The equation of tangent to the parabola $y^{2} = 6 x$ at point $(6, 6)$ is

2 x + 3 y - 30 = 0

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2 x - y - 6 = 0

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2 y - x - 6 = 0

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x + y - 12 = 0

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Solution

The correct option is C $2 y - x - 6 = 0$
Given parabola,
$y^{2} = 6 x$
The equation of tangent at $(6, 6)$ is
$T = 0 \Rightarrow y \cdot 6 = 6 (\frac{x + 6}{2}) \Rightarrow 12 y = 6 x + 36$

Hence, the required equation of tangent is
$2 y - x - 6 = 0$

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Perpendicular Distance of a Point from a Line

Standard XII Mathematics

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The equation of tangent to the parabola y2=6x at point (6,6) is

The correct option is C 2y−x−6=0Given parabola, y2=6x The equation of tangent at (6,6) is T=0⇒y⋅6=6(x+62)⇒12y=6x+36 Hence, the required equation of tangent is 2y−x−6=0

The equation of tangent to the parabola $y^{2} = 6 x$ at point $(6, 6)$ is

The correct option is C $2 y - x - 6 = 0$
Given parabola,
$y^{2} = 6 x$
The equation of tangent at $(6, 6)$ is
$T = 0 \Rightarrow y \cdot 6 = 6 (\frac{x + 6}{2}) \Rightarrow 12 y = 6 x + 36$

Hence, the required equation of tangent is
$2 y - x - 6 = 0$