Let's solve each part of the problem step by step:

Part (a): Finding the coordinates of A and B

1. Intersection of $y = x$ with the circle $x^2 + y^2 = 1$:

   • Substitute $y = x$ into the equation of the circle: $x^2 + x^2 = 1$ $2x^2 = 1$ $x^2 = \frac{1}{2} \implies x = \frac{1}{\sqrt{2}} \text{ or } x = -\frac{1}{\sqrt{2}}$
   • Since we're interested in the positive intersection (as shown in the diagram), we choose $x = \frac{1}{\sqrt{2}}$. Then, $y = \frac{1}{\sqrt{2}}$.
   • So, the coordinates of $A$ are: $A = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

2. Intersection of $y = x$ with the parabola $y = x^2$:

   • Substitute $y = x$ into $y = x^2$: $x = x^2$ $x^2 - x = 0$ $x(x - 1) = 0$
   • This gives $x = 0$ or $x = 1$. Since we are looking for the non-zero intersection (as shown in the diagram), we choose $x = 1$. Then, $y = 1$.
   • So, the coordinates of $B$ are: $B = (1, 1)$

Part (b): Finding the area of triangles $OAX$ and $OBY$

1. Triangle $OAX$:

   • The vertices are $O = (0, 0)$, $A = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, and $X = \left(\frac{1}{\sqrt{2}}, 0\right)$.
   • The base of triangle $OAX$ is $\frac{1}{\sqrt{2}}$ (the horizontal distance from $O$ to $X$).
   • The height of triangle $OAX$ is $\frac{1}{\sqrt{2}}$ (the vertical distance from $X$ to $A$).
   • The area of $OAX$ is: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{4}$

2. Triangle $OBY$:

   • The vertices are $O = (0, 0)$, $B = (1, 1)$, and $Y = (1, 0)$.
   • The base of triangle $OBY$ is $1$ (the horizontal distance from $O$ to $Y$).
   • The height of triangle $OBY$ is $1$ (the vertical distance from $Y$ to $B$).
   • The area of $OBY$ is: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$

Summary of Answers

• Coordinates:
  • $A = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
  • $B = (1, 1)$
• Areas:
  • Area of $\triangle OAX = \frac{1}{4}$
  • Area of $\triangle OBY = \frac{1}{2}$

Would you like further details or explanations on any part of the solution?

Follow-up Questions:

1. How would the coordinates of $A$ and $B$ change if the line was $y = -x$?
2. Can you verify the calculation of the areas using a different method, such as using vectors?
3. What would be the area of the triangle if the parabola were $y = x^3$ instead?
4. How would the solution change if the circle's equation was $x^2 + y^2 = 4$?
5. What if the line intersected the parabola at a different angle (e.g., $y = 2x$)?

Tip:

Remember that when working with intersections of curves, substituting one equation into another is a common method to find points of intersection. Always check which solution fits the context (e.g., positive or negative intersections).