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Locus of the midpoint of any normal chord of \mathrm{y^{2}=4 a x} is

Option: 1

\mathrm{x=a\left(\frac{4 a^{2}}{y^{2}}-2+\frac{y^{2}}{2 a^{2}}\right)}


Option: 2

\mathrm{x=a\left(\frac{4 a^2}{y^2}+2+\frac{y^2}{2 a^2}\right)}


Option: 3

\mathrm{x=a\left(\frac{4 a^2}{y^2}-2-\frac{y^2}{2 a^2}\right) }


Option: 4

\mathrm{x=a\left(\frac{4 a^2}{y^2}+2-\frac{y^2}{2 a^2}\right)}


Answers (1)

best_answer

Let \mathrm{A B}  be a normal chord where

\mathrm{A \equiv\left(a^{t_{1}^{2}}, 2^{t_{1}}\right), B \equiv\left(a^{t_{2}^{2}}, 2 t_{2}\right)}
If it's midpoint is \mathrm{P(h, k)}, then

\mathrm{2 h=a\left(t_{1}^{2}+t_{2}^{2}\right) }
       \mathrm{=a\left[\left(t_{1}+t_{2}\right)^{2}-2 t_{1} t_{2}\right] }

and \mathrm{2 k=2 a\left(t_{1}+t_{2}\right) }

We also have \mathrm{t_{2}=-t_{1}-\frac{2}{t_{1}}}

\mathrm{\mathrm{t}_{1}=-\frac{2 \mathrm{a}}{\mathrm{k}}\: and\: \mathrm{h}=\mathrm{a}\left(\mathrm{t}_{1}^{2}+2+\frac{2}{\mathrm{t}_{1}^{2}}\right)}
Thus required locus is
\mathrm{x=a\left(\frac{4 a^{2}}{y^{2}}+2+\frac{y^{2}}{2 a^{2}}\right)}.

Posted by

SANGALDEEP SINGH

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