#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 14 Maths Textbook Solution.

ANSWER: Equation of tangent, $x+y-2=0$

Equation  of normal , $y-x=0$

HINTS:

Differentiating the given curve with respect to x and find its slope first.

GIVEN:

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=2 \text { at }(1,1)$

SOLUTION:

\begin{aligned} &\frac{2}{3} x^{\frac{-1}{3}}+\frac{2}{3} y^{\frac{-1}{3}} \frac{d y}{d x}=0 \\ &\Rightarrow \frac{d y}{d x}=-\frac{x^{\frac{-1}{3}}}{y^{\frac{-1}{3}}}=\frac{-y^{\frac{1}{3}}}{x^{\frac{1}{3}}} \end{aligned}

Slope of tangent,

$m=\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{-1}{1}=-1$

Equation of tangent is

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-1=-1(x-1) \\ &\Rightarrow y-1=-x+1 \\ &\therefore x+y-2=0 \end{aligned}

Equation of Normal  is ,

\begin{aligned} &y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-1=1(x-1) \\ &\Rightarrow y-1=x-1 \\ &\therefore y-x=0 \end{aligned}