Answer to Question #320682 in Calculus for Aine Ndeutenge

Question

Show that the length of the portion of any tangent line to the asteroid a𝑥^2/3 + 𝑦^2/3 = 𝑎^2/3 ,cut off by the coordinate axes is constant.

Answer 1

$x^{2/3}+y^{2/3}=a^{2/3}\\\frac{2}{3}x^{-1/3}dx+\frac{2}{3}y^{-1/3}dy=0\Rightarrow \frac{dy}{dx}=-\frac{y^{1/3}}{x^{1/3}}\\The\,\,\tan gent\,\,line\,\,at\,\,\left( x_0,y_0 \right) :\\y-y_0=-\frac{{y_0}^{1/3}}{{x_0}^{1/3}}\left( x-x_0 \right) \\The\,\,intercept\,\,with\,\,Ox:\\y=0\Rightarrow x=x_0+{x_0}^{1/3}{y_0}^{2/3}\\The\,\,intercept\,\,with\,\,Oy:\\x=0\Rightarrow y=y_0+{x_0}^{2/3}{y_0}^{1/3}\\The\,\,dis\tan ce\,\,between\,\,\left( x_0+{x_0}^{1/3}{y_0}^{2/3},0 \right) ,\left( 0,y_0+{x_0}^{2/3}{y_0}^{1/3} \right) :\\d=\sqrt{\left( x_0+{x_0}^{1/3}{y_0}^{2/3} \right) ^2+\left( y_0+{x_0}^{2/3}{y_0}^{1/3} \right) ^2}=\\=\sqrt{{x_0}^{2/3}\left( {x_0}^{2/3}+{y_0}^{2/3} \right) ^2+{y_0}^{2/3}\left( {x_0}^{2/3}+{y_0}^{2/3} \right) ^2}=\left( {x_0}^{2/3}+{y_0}^{2/3} \right) ^{3/2}=a$