First, we'll let the point on the curve be called #M(x_0,y_0)#, 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#
#4x^3+4y^3*dy/dx=0#
#4y^3*dy/dx=-4x^3#
#dy/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(x-x_0)#
#yy_0^3-y_0^4=-x_0^3x+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_0^3p+x_0^4#
#x_0^3p=x_0^4+y_0^4#
#p=(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^(-4/3) +q^(-4/3)#
#=(a^4/x_0^3)^(-4/3) +(a^4/y_0^3)^(-4/3)#
#=(x_0^3/a^4)^(4/3) +(y_0^3/a^4)^(4/3)#
#=(x_0^4/a^(16/3)) +(y_0^4/a^(16/3))#
#=(x_0^4+y_0^4)/a^(16/3)#
#=(a^4)/a^(16/3)#
#=a^(-4/3)#
