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  • The equation of the tangent to the parabola  , which is perpendicular to

The equation of the tangent to the parabola  \mathrm{y^2=16 x}, which is perpendicular to the line \mathrm{y=3 x+7} is

Option: 1

\mathrm{y-3 x+4=0}


Option: 2

\mathrm{3 y-x+36=0}


Option: 3

\mathrm{3 y+x-36=0}


Option: 4

\mathrm{3 y+x+36=0}


Answers (1)

best_answer

A line perpendicular to the given line is \mathrm{ 3 y+x=\lambda \Rightarrow y=-\frac{1}{3} x+\frac{\lambda}{3}}

Here \mathrm{ m=-\frac{1}{3}, \quad c=\frac{\lambda}{3}} . If we compare \mathrm{ y^2=16 x} with \mathrm{ y^2=4 a x} , then \mathrm{ a=4}

Condition for tangency is \mathrm{ \quad c=\frac{a}{m} \Rightarrow \frac{\lambda}{3}=\frac{4}{(-1 / 3)} \Rightarrow \lambda=-36 . \, \, \, \, \, \, \therefore \quad} Required equation is \mathrm{x+3 y+36=0}

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jitender.kumar

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