Question

If the points of intersection of two distinct conics $x^2 + y^2 = 4b$ and $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ lie on the curve $y^2 = 3x^2$ then $3\sqrt{3}$ times the area of the rectangle formed by the intersection points is __.

Answer 1

Step 1: Substitute $y^2 = 3x^2$ in Both Conics

From $y^2 = 3x^2$, substitute into $x^2 + y^2 = 4b$ and $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ to find values of $b$.

Step 2: Solve for $b$

By substituting, we get:

$x^2 = b \quad \text{and} \quad \frac{b}{16} + \frac{3}{b} = 1$

Solving this equation gives $b = 4$ or $b = 12$. Since $b = 4$ makes the curves coincide, we reject it, so $b = 12$.

Step 3: Find Points of Intersection

With $b = 12$, the points of intersection are $\left(\pm \sqrt{12}, \pm 6\right)$.

Step 4: Calculate the Area of the Rectangle

The area of the rectangle formed by these points is:

$\text{Area} = 2 \cdot \sqrt{12} \times 2 \cdot 6 = 4 \cdot \sqrt{12} \cdot 6 = 432$

So, the correct answer is: 432