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

Jan 10, 2018

$= {a}^{- \frac{4}{3}}$

#### Explanation:

First, we'll let the point on the curve be called $M \left({x}_{0} , {y}_{0}\right)$, keeping in mind that ${x}_{0}^{4} + {y}_{0}^{4} = {a}^{4}$.

Then to find an equation for the tangent, we'll need the gradient at point M. We COULD subject $y$ and derive, but you'll see it gets kinda messy. For this we'll use implicit derivation with respect to $x$:

${x}^{4} + {y}^{4} = {a}^{4}$
$4 {x}^{3} + 4 {y}^{3} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$4 {y}^{3} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - 4 {x}^{3}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{3} / {y}^{3}$
So the point M, the slope is $- {x}_{0}^{3} / {y}_{0}^{3}$

We use point-gradient formula to find an equation for the tangent:
$y - {y}_{0} = - {x}_{0}^{3} / {y}_{0}^{3} \left(x - {x}_{0}\right)$
$y {y}_{0}^{3} - {y}_{0}^{4} = - {x}_{0}^{3} x + {x}_{0}^{4}$

We let $p$ be the x-intercept, and $q$ be the y-intercept:
At x-int, y = 0:
$- {y}_{0}^{4} = - {x}_{0}^{3} p + {x}_{0}^{4}$
${x}_{0}^{3} p = {x}_{0}^{4} + {y}_{0}^{4}$
$p = \frac{{x}_{0}^{4} + {y}_{0}^{4}}{{x}_{0}^{3}}$, but remember the observation we made at the start, so:
$p = {a}^{4} / {x}_{0}^{3}$

Similarly, $q = {a}^{4} / {y}_{0}^{3}$

Now we evaluate ${p}^{- \frac{4}{3}} + {q}^{- \frac{4}{3}}$
$= {\left({a}^{4} / {x}_{0}^{3}\right)}^{- \frac{4}{3}} + {\left({a}^{4} / {y}_{0}^{3}\right)}^{- \frac{4}{3}}$
$= {\left({x}_{0}^{3} / {a}^{4}\right)}^{\frac{4}{3}} + {\left({y}_{0}^{3} / {a}^{4}\right)}^{\frac{4}{3}}$
$= \left({x}_{0}^{4} / {a}^{\frac{16}{3}}\right) + \left({y}_{0}^{4} / {a}^{\frac{16}{3}}\right)$
$= \frac{{x}_{0}^{4} + {y}_{0}^{4}}{a} ^ \left(\frac{16}{3}\right)$
$= \frac{{a}^{4}}{a} ^ \left(\frac{16}{3}\right)$
$= {a}^{- \frac{4}{3}}$