Key Concepts and Formulas

• Intersection of Curves: To find the points of intersection of two curves, solve their equations simultaneously.
• Substitution: If the intersection points of two curves lie on a third curve, substitute the third curve's equation into the equations of the first two curves.
• Area of a Rectangle: The area of a rectangle with sides of length $l$ and $w$ is $A = lw$.

Step-by-Step Solution

Step 1: State the given equations and constraint.

We are given the following equations: $x^2 + y^2 = 4b \quad (1)$ $\frac{x^2}{16} + \frac{y^2}{b^2} = 1 \quad (2)$ $y^2 = 3x^2 \quad (3)$

The goal is to find the value of $3\sqrt{3}$ times the area of the rectangle formed by the intersection points of (1) and (2), given that these points lie on (3).

Step 2: Substitute the constraint $y^2 = 3x^2$ into equation (1).

Substituting $y^2 = 3x^2$ into $x^2 + y^2 = 4b$, we get: $x^2 + 3x^2 = 4b$ $4x^2 = 4b$ $x^2 = b \quad (4)$

Step 3: Substitute the constraint $y^2 = 3x^2$ into equation (2).

Substituting $y^2 = 3x^2$ into $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$, we get: $\frac{x^2}{16} + \frac{3x^2}{b^2} = 1 \quad (5)$

Step 4: Substitute $x^2 = b$ from equation (4) into equation (5).

Substituting $x^2 = b$ into $\frac{x^2}{16} + \frac{3x^2}{b^2} = 1$, we get: $\frac{b}{16} + \frac{3b}{b^2} = 1$ $\frac{b}{16} + \frac{3}{b} = 1$

Step 5: Solve for $b$.

Multiplying both sides of the equation by $16b$, we get: $b^2 + 48 = 16b$ $b^2 - 16b + 48 = 0$ Factoring the quadratic equation, we get: $(b - 4)(b - 12) = 0$ So, $b = 4$ or $b = 12$.

Step 6: Validate the values of $b$ using the "distinct conics" condition.

If $b = 4$, then equation (1) becomes $x^2 + y^2 = 16$, and equation (2) becomes $\frac{x^2}{16} + \frac{y^2}{16} = 1$, which simplifies to $x^2 + y^2 = 16$. Since the two equations are identical, the conics are not distinct. Therefore, $b = 4$ is not a valid solution.

If $b = 12$, then equation (1) becomes $x^2 + y^2 = 48$, and equation (2) becomes $\frac{x^2}{16} + \frac{y^2}{144} = 1$. These conics are distinct. Therefore, $b = 12$ is the valid solution.

Step 7: Find the coordinates of the intersection points.

Since $x^2 = b$ and $b = 12$, we have $x^2 = 12$, so $x = \pm 2\sqrt{3}$. Since $y^2 = 3x^2$, we have $y^2 = 3(12) = 36$, so $y = \pm 6$.

The four intersection points are $(2\sqrt{3}, 6)$, $(2\sqrt{3}, -6)$, $(-2\sqrt{3}, 6)$, and $(-2\sqrt{3}, -6)$.

Step 8: Calculate the area of the rectangle formed by the intersection points.

The vertices of the rectangle are $(2\sqrt{3}, 6)$, $(2\sqrt{3}, -6)$, $(-2\sqrt{3}, 6)$, and $(-2\sqrt{3}, -6)$. The length of the rectangle is $2(2\sqrt{3}) = 4\sqrt{3}$, and the width of the rectangle is $2(6) = 12$. The area of the rectangle is $(4\sqrt{3})(12) = 48\sqrt{3}$.

Step 9: Calculate $3\sqrt{3}$ times the area of the rectangle.

The required value is $3\sqrt{3} \times (48\sqrt{3}) = 3 \times 48 \times 3 = 144 \times 3 = 432$.

Common Mistakes & Tips

• Remember to consider both positive and negative roots when solving for $x$ and $y$.
• Always check the validity of your solutions, especially when the problem states a condition like "distinct conics".
• Carefully substitute the constraint equation to simplify the equations and solve for the unknown parameter.

Summary

We found the intersection points of two distinct conics by using the equation of a third curve as a constraint. We substituted the constraint into the equations of the conics, solved for the parameter $b$, and found the coordinates of the intersection points. We then calculated the area of the rectangle formed by these points and multiplied it by $3\sqrt{3}$ to get the final answer.

The final answer is \boxed{432}.