Explanation
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.
The abscissa of the point on the curve $$\sqrt {xy} = a + x$$ , the tangent at which cuts off equal intercepts from the co-ordinate axes is (a >0)
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