The equation of the normal to the parabola $x^{2} = 8 y$ whose slope is $\frac{1}{m}$ is

y = m x - 2 m - m^{3}

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x = m x - 4 m - {2 m}^{3}

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x = m y - 4 m - 2 m^{3}

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4 y = m x + 4 m - 2 m^{3}

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Solution

The correct option is C $x = m y - 4 m - 2 m^{3}$

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Similar questions

Given slope m, find the equation of normal having slope m to the parabola $y^{2} = 4 a x$

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 ?

x^{3} - y^{2} = 0

(m^{2}, - m^{3})

= - \frac{1}{m} x - 2 m^{3},

m^{2}

x^{2} - (2 + m) x + 1 (m^{2} - 4 m + 4) = 0

What's the equation of tangent to the parabola $y^{2} = 4 a x$ having a slope 'm'.

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