Application of Derivatives Test - 61

Consider the given expression.

$${{x}^{4}}+{{y}^{4}}={{a}^{4}}$$

 Let the point on the curve be $$M\left( {{x}_{0}},{{y}_{0}} \right)$$.

 Therefore,

$$x_{0}^{4}+x_{0}^{4}={{a}^{4}}$$

 Differentiate the expression given in the question with respect to $$x$$.

$$ {{x}^{4}}+{{y}^{4}}={{a}^{4}} $$

$$ 4{{x}^{3}}+4{{y}^{3}}\dfrac{dy}{dx}=0 $$

$$ \dfrac{dy}{dx}=-\dfrac{{{x}^{3}}}{{{y}^{3}}} $$

 So, at the point $$M$$, the slope is,

$$\Rightarrow -\dfrac{x_{0}^{3}}{y_{0}^{3}}$$

 Therefore, equation of the tangent is,

$$\begin{align}

$$ y-{{y}_{0}}=-\dfrac{x_{0}^{3}}{y_{0}^{3}}\left( x-{{x}_{0}} \right) $$

$$ yy_{0}^{3}-y_{0}^{4}=-x_{0}^{3}x+x_{0}^{4} $$

\end{align}$$

 Let $$p$$ and $$q$$ be the $$x$$ and $$y$$ intercept, respectively. So, at $$x$$ intercept,

$$ -y_{0}^{4}=-x_{0}^{3}p+x_{0}^{4} $$

$$ x_{0}^{3}p=x_{0}^{4}+y_{0}^{4} $$

$$ p=\dfrac{x_{0}^{4}+y_{0}^{4}}{x_{0}^{3}} $$

$$ p=\dfrac{{{a}^{4}}}{x_{0}^{3}} $$

 Similarly,

$$q=\dfrac{{{a}^{4}}}{y_{0}^{3}}$$

 Now, calculate the value of $${{p}^{-4/3}}+{{q}^{-4/3}}$$.

$$ ={{\left( \dfrac{{{a}^{4}}}{x_{0}^{3}} \right)}^{-4/3}}+{{\left( \dfrac{{{a}^{4}}}{y_{0}^{3}} \right)}^{-4/3}} $$

$$ ={{\left( \dfrac{x_{0}^{3}}{{{a}^{4}}} \right)}^{4/3}}+{{\left( \dfrac{y_{0}^{3}}{{{a}^{4}}} \right)}^{4/3}} $$

$$ =\left( \dfrac{x_{0}^{4}}{{{a}^{16/3}}} \right)+\left( \dfrac{y_{0}^{4}}{{{a}^{16/3}}} \right) $$

$$ =\dfrac{x_{0}^{4}+y_{0}^{4}}{{{a}^{16/3}}} $$

$$ =\dfrac{{{a}^{4}}}{{{a}^{16/3}}} $$

$$ ={{a}^{-4/3}} $$

 Hence, this is the required result.