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 __.
Correct Answer: 432
Approach Solution - 1
To solve the problem, we need to find the intersection points of the given conics and verify that they lie on the curve \(y^2 = 3x^2\). Analyze each step as follows:
- Conics: Solve \(x^2+y^2=4b\) and \(\frac{x^2}{16}+\frac{y^2}{b^2}=1\) simultaneously.
- From the second conic: \(b^2x^2+16y^2=16b^2\).
- Eliminate \(y^2\) by substituting from the first conic: \(x^2=4b-y^2\). Substitute into the second equation: \(b^2(4b-y^2)+16y^2=16b^2\).
- Rearrange: \(4b^3-b^2y^2+16y^2=16b^2\). Simplify to \(y^2(b^2+16)=4b^3-16b^2\). Express \(y^2=4b(1-b)\).
- Intersection with \(y^2=3x^2\): Substitute \(y^2=3x^2\) in \((x^2+y^2=4b)\), giving \(x^2+3x^2=4b\). Solve \(4x^2=4b \Rightarrow x^2=b\). Then, \(y^2=3b\).
- Calculate rectangle area: Vertices \((\sqrt{b}, \sqrt{3b}),(-\sqrt{b},-\sqrt{3b}),(-\sqrt{b},\sqrt{3b}), (\sqrt{b},-\sqrt{3b})\). Length = \(2\sqrt{b}\), width = \(2\sqrt{3b}\). Thus, Area = \(4b\sqrt{3}\).
- Multiply area by \(3\sqrt{3}\): \((3\sqrt{3}) \times (4b\sqrt{3}) = 36b\).
- Given range \(432,432\), solve \(36b=432\) to find \(b=12\).
- Finally, area calculation \((36b)\) at \(b=12\) is 432, fitting the range 432.
The final product, \(3\sqrt{3}\) times the area, is 432.
Approach Solution -2
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
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