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  • The locus of the poles of the normal chard of the parabola is  </

The locus of the poles of the normal chard of the parabola \mathrm{y^2=4 a x} is
 

Option: 1

\mathrm{y^2 x+4 a^3=0}
 


Option: 2

\mathrm{y^2(x+2 a)=0}
 


Option: 3

\mathrm{y(x+2 a)+4 a^3=0}
 


Option: 4

\mathrm{y^2(x+2 a)+4 a^3=0.}


Answers (1)

best_answer

Let the equation of the normal be

\mathrm{ y=m x-2 a m-a m^3 }  .......(1)

If \mathrm{ (h, k) } be its pole then its equation is

\mathrm{ k y=2 a(x+h) }

\mathrm{ \text { or } y=\frac{2 a}{k} x+2 a \frac{h}{k} }         .........(2)

Comparing (1) and (2), we get

\mathrm{ m=\frac{2 a}{k},-2 a m-a m^3=2 a \cdot \frac{h}{k} }

Eliminating the variable \mathrm{ m } between the above two relations we get

\mathrm{ -2 a \cdot\left(\frac{2 a}{k}\right)-a \cdot \frac{8 a^3}{k^3}=2 a \cdot \frac{h}{k} }

\mathrm{ \text { or } h+2 a+\frac{4 a^3}{k^2}=0 \text { or } k^2(h+2 a)+4 a^3=0 }

Hence the required locus is

\mathrm{ y^2(x+2 a)+4 a^3=0 }
 

Posted by

Ritika Harsh

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