Question

If the tangent at any point on the curve $x^4+y^4=a^4$ cuts off intercepts $p$ and $q$ on the coordinate axes the value of $p^{-4/3}+q^{-4/3}$ is

• A. $a^{-4/3}$
• B. $a^{-1/3}$
• C. $a^{1/2}$
• D. $Noneofthese$

Answer 1

The correct option is A $a^{-4/3}$

Consider the given expression.

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

Let the point on the curve be $M(x_0,y_0)$.

Therefore,

$x_0^4+x_0^4=a^4$

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

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

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

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

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

$\Rightarrow -\frac{x_0^3}{y_0^3}$

Therefore, equation of the tangent is,

$\begin{align}

$y-y_0=-\frac{x_0^3}{y_0^3}(x-x_0)$

$y{y}_0^3-y_0^4=-x_0^{3}x+x_0^4$

\end{align}$

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

$-y_0^4=-x_0^{3}p+x_0^4$

$x_0^{3}p=x_0^4+y_0^4$

$p=\frac{x_0^4+y_0^4}{x_0^3}$

$p=\frac{a^4}{x_0^3}$

Similarly,

$q=\frac{a^4}{y_0^3}$

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

$={\left(\frac{a^4}{x_0^3}\right)}^{-4/3}+{\left(\frac{a^4}{y_0^3}\right)}^{-4/3}$

$={\left(\frac{x_0^3}{a^4}\right)}^{4/3}+{\left(\frac{y_0^3}{a^4}\right)}^{4/3}$

$=\left(\frac{x_0^4}{a^{16/3}}\right)+\left(\frac{y_0^4}{a^{16/3}}\right)$

$=\frac{x_0^4+y_0^4}{a^{16/3}}$

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

$=a^{-4/3}$

Hence, this is the required result.