To analyze the curve described by the equation

$x^{\frac{p}{q}} + y^{\frac{p}{q}} = 1$

with the given condition that $\frac{p}{q} = \frac{4}{3}$, let’s follow these steps:

1. Substitute $\frac{p}{q} = \frac{4}{3}$ into the equation:

   $x^{\frac{4}{3}} + y^{\frac{4}{3}} = 1$

2. Implicit Differentiation: To find points where the slope of the tangent line is equal to 1, we need to find $\frac{dy}{dx}$ by differentiating both sides of the equation with respect to $x$.

   Differentiating $x^{\frac{4}{3}} + y^{\frac{4}{3}} = 1$ with respect to $x$:

   $\frac{4}{3}x^{\frac{4}{3} - 1} + \frac{4}{3}y^{\frac{4}{3} - 1} \cdot \frac{dy}{dx} = 0$

   Simplify this to:

   $\frac{4}{3}x^{\frac{1}{3}} + \frac{4}{3}y^{\frac{1}{3}} \cdot \frac{dy}{dx} = 0$

3. Solve for $\frac{dy}{dx}$:

   Rearranging terms,

   $\frac{dy}{dx} = -\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}}$

4. Set Slope Equal to 1: Now, we are asked to find the points where the slope of the tangent line is equal to 1. Therefore, set $\frac{dy}{dx} = 1$:

   $-\frac{x^{\frac{1}{3}}}{y^{\frac{1}{3}}} = 1$

   Solving this equation:

   $x^{\frac{1}{3}} = -y^{\frac{1}{3}}$

   Since $x^{\frac{4}{3}} + y^{\frac{4}{3}} = 1$ only has solutions for real values when $x$ and $y$ are positive, no real solution exists for points where the slope is exactly 1 in this scenario.

Let me know if you'd like more details on each step or have any questions. Here are some related questions to explore this topic further:

1. How does implicit differentiation work with more complex exponents?
2. What happens if $\frac{p}{q}$ has a different rational value?
3. How can we interpret solutions in the complex plane?
4. What other curves have similar symmetrical properties?
5. How do we find normal lines instead of tangents for such curves?

Tip: When dealing with equations involving rational exponents, be cautious of sign constraints that limit real solutions.