If A = {x, y) | x2 + y2 = 25} and B = {x, y) | x 2 + 9y2 = 144} then BA ∩ contains :

1. Break Down the Problem

We need to analyze the sets A and B defined by their respective equations:

• Set A is the equation of a circle.
• Set B is the equation of an ellipse.

We need to find the points where the circle intersects with the ellipse, expressed as $B \cap A$.

2. Relevant Concepts

• The equation of the circle $A$: $x^2 + y^2 = 25$ This is a circle centered at the origin (0, 0) with a radius of 5.

• The equation of the ellipse $B$: $x^2 + 9y^2 = 144$ This can be rearranged to: $\frac{x^2}{144} + \frac{y^2}{16} = 1$ This represents an ellipse centered at the origin with semi-major axis 12 (along the x-axis) and semi-minor axis 4 (along the y-axis).

3. Analysis and Detail

Step 1: Solve for $y^2$ from the circle's equation

From the circle equation, we have: $y^2 = 25 - x^2$

Step 2: Substitute $y^2$ into the ellipse's equation

Substituting $y^2$ into the ellipse equation: $x^2 + 9(25 - x^2) = 144$ This simplifies to: $x^2 + 225 - 9x^2 = 144$ $-8x^2 + 225 = 144 \implies -8x^2 = 144 - 225 \implies -8x^2 = -81 \implies x^2 = \frac{81}{8}$ $x = \pm \sqrt{\frac{81}{8}} = \pm \frac{9}{\sqrt{8}} = \pm \frac{9\sqrt{2}}{4}$

Step 3: Solve for $y$ using $x$

Substituting $x$ back into the circle's equation to find $y$:

For $x = \frac{9\sqrt{2}}{4}$: $y^2 = 25 - \left(\frac{9\sqrt{2}}{4}\right)^2 = 25 - \frac{81 \cdot 2}{16} = 25 - \frac{162}{16} = \frac{400 - 162}{16} = \frac{238}{16} = \frac{119}{8}$ $y = \pm \sqrt{\frac{119}{8}} = \pm \frac{\sqrt{119}}{2\sqrt{2}} = \pm \frac{\sqrt{238}}{4}$

4. Verify and Summarize

The intersection points are given as: $\left( \frac{9\sqrt{2}}{4}, \frac{\sqrt{238}}{4} \right) \quad \text{and} \quad \left( \frac{9\sqrt{2}}{4}, -\frac{\sqrt{238}}{4} \right)$ $\left( -\frac{9\sqrt{2}}{4}, \frac{\sqrt{238}}{4} \right) \quad \text{and} \quad \left( -\frac{9\sqrt{2}}{4}, -\frac{\sqrt{238}}{4} \right)$

Final Answer

The intersection $B \cap A$ contains the points:

1. $\left( \frac{9\sqrt{2}}{4}, \frac{\sqrt{238}}{4} \right)$
2. $\left( \frac{9\sqrt{2}}{4}, -\frac{\sqrt{238}}{4} \right)$
3. $\left( -\frac{9\sqrt{2}}{4}, \frac{\sqrt{238}}{4} \right)$
4. $\left( -\frac{9\sqrt{2}}{4}, -\frac{\sqrt{238}}{4} \right)$