Question

x^2/3+y^2/3=2 at(1,1) find the equation of the tangent and normal line of the following curve

Answer 1

Dear student
$$\begin{gathered} Thegivenequationis{x}^{\frac{2}{3}}+y^{\frac{2}{3}}=2 \\ Ondifferentiatingw.r.t.x,weget \\ \Rightarrow \frac{2}{3}{x}^{\frac{-1}{3}}+\frac{2}{3}{y}^{\frac{-1}{3}}\times \frac{d}{dx}\left(y\right)=0 \\ \Rightarrow \frac{2}{3}\left[x^{\frac{-1}{3}}+y^{\frac{-1}{3}}\times \frac{dy}{dx}\right]=0 \\ \Rightarrow \left[x^{\frac{-1}{3}}+y^{\frac{-1}{3}}\times \frac{dy}{dx}\right]=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{x^{\frac{-1}{3}}}{y^{\frac{-1}{3}}} \\ \therefore Theslopeofthe\tan gentat\left(1,1\right)is{\frac{dy}{dx}}_{at\left(1,1\right)}=1 \\ Then,theequationofthe\tan gentat\left(1,1\right)isgivenby \\ y-1=1\left(x-1\right) \\ \Rightarrow x-y=0 \\ Now,theslopeofthenormalat\left(1,1\right)isgivenby \\ \frac{-1}{Slopeofthe\tan gentat\left(1,1\right)}=-1 \\ Thus,theequationofthenormalat\left(1,1\right)isgivenas: \\ y-1=-1\left(x-1\right) \\ \Rightarrow y-1=-x+1 \\ \Rightarrow x+y=2 \end{gathered}$$
Regards