Find the equation of normal to the parabola $y^{2} = 6 x,$ at the point whose ordinate is $6$ .

2 x + y - 18 = 0

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2 x + y - 20 = 0

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x + 2 y - 18 = 0

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2 x + 4 y - 18 = 0

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Solution

The correct option is A $2 x + y - 18 = 0$
Given equation of parabola is $y^{2} = 6 x$
$∴ 2 y \frac{d y}{d x} = 6$
Given $y$ -coordinate $= 6$
$\Rightarrow x = 6$
So, the point is $(6, 6)$
Slope of tangent to the parabola at $(6, 6)$ is $\frac{1}{2}$
So, slope of normal is -2.
Equation of normal at $(6, 6)$ is
$y - 6 = - 2 (x - 6)$
$\Rightarrow y - 6 = - 2 x + 12$
$\Rightarrow y + 2 x - 18 = 0$

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Find the equation of normal to the parabola y2=6x, at the point whose ordinate is 6.

The correct option is A 2x+y−18=0Given equation of parabola is y2=6x∴2ydydx=6Given y-coordinate =6⇒x=6So, the point is (6,6)Slope of tangent to the parabola at (6,6) is 12So, slope of normal is -2.Equation of normal at (6,6) isy−6=−2(x−6)⇒y−6=−2x+12⇒y+2x−18=0

Find the equation of normal to the parabola $y^{2} = 6 x,$ at the point whose ordinate is $6$ .