To find the foci and vertices of the given equation $y - x = \frac{1}{y + x}$, we must first manipulate it into a form that we can recognize as an ellipse, hyperbola, or parabola equation.

Step 1: Substitute variables

Let: $u = y - x \quad \text{and} \quad v = y + x.$ Then the equation becomes: $u = \frac{1}{v}.$

Step 2: Rewrite the equation

Multiplying both sides by $v$ to eliminate the fraction: $uv = 1.$ This is a hyperbolic equation. It can be recognized as a hyperbola of the form $xy = 1$, but in terms of the new variables $u$ and $v$.

Step 3: Transform back to $x$ and $y$

Now, reverse the substitutions: $u = y - x \quad \text{and} \quad v = y + x.$ So the equation becomes: $(y - x)(y + x) = 1.$ Using the difference of squares formula: $y^2 - x^2 = 1.$ This is the equation of a hyperbola.

Step 4: Standard form of the hyperbola

The equation $y^2 - x^2 = 1$ is already in the standard form of a hyperbola: $\frac{y^2}{1} - \frac{x^2}{1} = 1,$ which represents a hyperbola centered at $(0, 0)$, opening along the $y$-axis.

Step 5: Find the vertices and foci

• Vertices: The vertices of the hyperbola are at $(0, \pm 1)$, because the equation involves $y^2$ with a coefficient of 1.

• Foci: The foci are found using the formula $c^2 = a^2 + b^2$, where $a = 1$ and $b = 1$ (since the denominators in the hyperbola's standard form are both 1).

  Therefore: $c^2 = 1 + 1 = 2 \quad \Rightarrow \quad c = \sqrt{2}.$ The foci are located at $(0, \pm \sqrt{2})$.

Final Answer:

• Vertices: $(0, 1)$ and $(0, -1)$.
• Foci: $(0, \sqrt{2})$ and $(0, -\sqrt{2})$.

Would you like further details or explanations? Here are some additional questions you could explore:

1. How is the equation for a hyperbola derived from its geometric definition?
2. Can you find the asymptotes of the hyperbola $y^2 - x^2 = 1$?
3. What are the parametric equations for this hyperbola?
4. How do you differentiate between equations of ellipses and hyperbolas?
5. Can you solve for the eccentricity of this hyperbola?

Tip: The foci of a hyperbola are always further from the center than the vertices, as the distance between them is determined by $c = \sqrt{a^2 + b^2}$.