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Tangent to curve $x^3+y^3=a^3$ at $(x_1,y_1)$ meets it again at $(x_2,y_2)$.How to prove that $$\frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0$$


Since $y'=-\frac{x_1^2}{y_1^2}$ $$\frac{y_2-y_1}{x_2-x_1}=-\frac{x_1^2}{y_1^2}=\frac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$$ Solving we get: $$\frac{x_2}{x_1}+\frac{y_2}{y_1}+\frac{x_2^2}{x_1^2}+\frac{y_2^2}{y_1^2}+2=0$$ Or $$\left(\frac{x_2}{x_1}+\frac{y_2}{y_1}\right)+\left(\frac{x_2^2}{x_1^2}+\frac{y_2^2}{y_1^2}\right)^2=2\frac{x_2y_2-x_1y_1}{x_1y_1}$$ Which isn't what is nedded to prove. can someone guide me?

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$$\frac{y_2-y_1}{x_2-x_1} = \frac{y_2^3-y_1^3}{x_2^3-x_1} \times \frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2} = (-)\frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2}$$ And you have written $$\frac{y_2-y_1}{x_2-x_1} = \left(\frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2}\right) $$ Which is wrong.


So we have $$\frac{x_1^2}{y_1^2} = \frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2} $$

This gives, $$ x_1^2y_2^2+x_1^2y_1y_2+x_1^2y_1^2=y_1^2x_1x_2+x_1^2y_1^2+x_2^2y_1^2 $$ $$ x_1^2y_2^2+x_1^2y_1y_2=y_1^2x_1x_2+x_2^2y_1^2 $$ $$(x_1y_2+x_2y_1)(x_1y_2-x_2y_1)= x_1y_1(x_2y_1-y_2x_1)$$ $$ \frac{x_2}{x_1}+\frac{y_2}{y_1}+1 = 0 \Box $$

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  • $\begingroup$ Better use $$\frac{x_1^2}{y_1^2} = \frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2}$$$$=\frac{x_2^2+x_1x_2}{y_2^2+y_1y_2}$$ $\endgroup$ – RE60K Aug 14 '14 at 12:46
  • $\begingroup$ Doesn't matter :), $\endgroup$ – Shivang jindal Aug 14 '14 at 12:48
  • $\begingroup$ You have to cross multiply anyway $\endgroup$ – Shivang jindal Aug 14 '14 at 12:49
  • $\begingroup$ Ok I followed your comment anyways $\endgroup$ – RE60K Aug 14 '14 at 12:49

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