Problem : If the tangent at $(x_0,y_0)$ to the curve $x^3+y^3=a^3$ meets the curve again at $(x_1,x_1)$ then prove that $\frac{x_0}{x_1}+\frac{y_0}{y_1}=1$
We have tangent to the given curve at $(x_0,y_0)$ is
$$xx_0^2+yy_0^2=a^3$$.
Since it passes through $(x_1,y_1)$, so we get $$x_1x_0^2+y_1y_0^2=a^3\tag{1}$$ Again, $(x_1,y_1)$ lies on the curve, so.
$$x_1^3+y_1^3=a^3\tag{2}$$.
How can I get the required result from above?
$$x^3+y^3=a^3$$ $$3x^2+3y^2 y'=0$$ $$y'=-\frac{x^2}{y^2}$$
Thus tangent at $(x_0,y_0) $ can be written as
$$(y-y_0)=-\frac{x_0^2}{y_0^2}(x-x_0)$$ $$yy_0^2=-x_0^2x+x_o^3+y_0^3$$
The intersection of this line and curve $x^3+y^3=a^3$ is at the point $(x_1,y_1)$
thus $(x_1,y_1)$ satiesfies both the line and the curve
Thus we can write $$x_1^3+y_1^3=a^3$$
and $$y_1y_0^2=-x_0^2x_1+x_0^3+y_0^3$$ $$y_1y_0^2+x_0^2x_1=x_0^3+y_0^3$$
also $x_0^3+y_0^3=a^3$ Thus $$y_1y_0^2+x_0^2x_1=a^3$$ $$\frac{y_1}{y_0}y_0^3+\frac{x_1}{x_0}x_0^3=a^3$$
We have $$y=(a^3-x^3)^{1/3}\implies y'=-\frac{x^2}{y^2}$$ so the tangent line is $$y-y_0=-\frac{x_0^2}{y_0^2}(x-x_0)\implies y_0^2y=-x_0^2x+a^3$$ Now at $(x_1,y_1)$, we get $$x_0^2x_1+y_0^2y_1=a^3\implies\left(\frac{x_0}{x_1}\right)^2x_1^3+\left(\frac{y_0}{y_1}\right)^2y_1^3=a^3$$ from which the result follows.
