Question #320682

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.



1
Expert's answer
2022-04-01T05:04:04-0400

x2/3+y2/3=a2/323x1/3dx+23y1/3dy=0dydx=y1/3x1/3Thetangentlineat(x0,y0):yy0=y01/3x01/3(xx0)TheinterceptwithOx:y=0x=x0+x01/3y02/3TheinterceptwithOy:x=0y=y0+x02/3y01/3Thedistancebetween(x0+x01/3y02/3,0),(0,y0+x02/3y01/3):d=(x0+x01/3y02/3)2+(y0+x02/3y01/3)2==x02/3(x02/3+y02/3)2+y02/3(x02/3+y02/3)2=(x02/3+y02/3)3/2=ax^{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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS