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

Correct Answer: 432

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