\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
By using the chain rule, we have \begin {align*} \DiffFrac {y}{x} & = \frac {\Diff y / \Diff t}{\Diff x / \Diff t} \\ & = \frac {12 \cos t - 12 \sin ^2 t \cos t}{12 \cos ^2 t \sin t} \\ & = \frac {\cos t - \sin ^2 t \cos t}{\cos ^2 t \sin t} \\ & = \frac {1 - \sin ^2 t}{\cos t \sin t} \\ & = \frac {\cos ^2 t}{\cos t \sin t} \\ & = \frac {\cos t}{\sin t} \\ & = \cot t. \end {align*}
Hence, at \(t = \phi \), the normal of this curve has gradient \(- \tan \phi \), and hence it has equation \begin {align*} y - \left (12 \sin \phi - 4 \sin ^3 \phi \right ) & = - \tan \phi \left (x - \left (-4 \cos ^3 \phi \right )\right ) \\ y - 12 \sin \phi + 4 \sin ^3 \phi & = - \tan \phi x - 4 \cos ^3 \phi \tan \phi \\ \cos \phi y - 12 \sin \phi \cos \phi + 4 \sin ^3 \phi \cos \phi & = - \sin \phi x - 4 \cos ^3 \phi \sin \phi \\ \sin \phi x + \cos \phi y & = 12 \sin \phi \cos \phi - 4 \sin ^3 \phi \cos \phi - 4 \cos ^3 \phi \sin \phi \\ \sin \phi x + \cos \phi y & = 4 \sin \phi \cos \phi \left (3 - \sin ^2 \phi - \cos ^2 \phi \right ) \\ \sin \phi x + \cos \phi y & = 8 \sin \phi \cos \phi . \end {align*}
The curve \(x^{\frac {2}{3}} + y^{\frac {2}{3}} = 4\) can be parametrised as \(x = 8 \cos ^3 t\) and \(y = 8 \sin ^3 t\): \begin {align*} x^{\frac {2}{3}} + y^{\frac {2}{3}} & = \left (8 \cos ^3 t\right )^{\frac {2}{3}} + \left (8 \sin ^3 t\right )^{\frac {2}{3}} \\ & = 4 \cos ^2 t + 4 \sin ^2 t \\ & = 4. \end {align*}
Hence, the gradient of the tangent at a point is \begin {align*} \DiffFrac {y}{x} & = \frac {\Diff y / \Diff t}{\Diff x / \Diff t} \\ & = \frac {24 \sin ^2 t \cos t}{-24 \cos ^2 t \sin t} \\ & = - \tan t, \end {align*}
and the equation of the tangent at the point \(t = \phi \) is \begin {align*} y - 8 \sin ^3 \phi & = - \tan \phi \left (x - 8 \cos ^3 \phi \right ) \\ \cos \phi y - 8 \sin ^3 \phi \cos \phi & = - \sin \phi x + 8 \cos ^3 \phi \sin \phi \\ \sin \phi x + \cos \phi y & = 8 \sin \phi \cos \phi \left (\sin ^2 \phi + \cos ^2 \phi \right ) \\ \sin \phi x + \cos \phi y & = 8 \sin \phi \cos \phi , \end {align*}
which shows the normal to the original curve is the tangent to this new curve at \((8 \cos ^3 \phi , 8 \sin ^3 \phi )\).
By using the chain rule, we have \begin {align*} \DiffFrac {y}{x} & = \frac {\Diff y / \Diff t}{\Diff x / \Diff t} \\ & = \frac {\cos t - \cos t + t \sin t}{- \sin t + \sin t + t \cos t} \\ & = \frac {t \sin t}{t \cos t} \\ & = \tan t. \end {align*}
Hence, at \(t = \phi \), the normal of this curve has gradient \(- \cot \phi \), and hence it has equation \begin {align*} y - \left (\sin \phi - \phi \cos \phi \right ) & = - \cot \phi \left (x - \left (\cos \phi + \phi \sin \phi \right )\right ) \\ \sin \phi y - \sin ^2 \phi + \phi \sin \phi \cos \phi & = - \cos \phi x + \cos ^2 \phi + \phi \sin \phi \cos \phi \\ \cos \phi x + \sin \phi y & = \sin ^2 \phi + \cos ^2 \phi \\ \cos \phi x + \sin \phi y & = 1. \end {align*}
The distance of this normal to the origin is \[ \frac {\abs *{\cos \phi \cdot 0 + \sin \phi \cdot 0 - 1}}{\sqrt {\cos ^2 \phi + \sin ^2 \phi }} = 1, \] which is a constant, and hence this curve is tangent to the unit circle \(x^2 + y^2 = 1\).