Let the ellipse $\mathbf{E}_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \mathbf{a} > \mathbf{b}$ and $\mathbf{E}_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, \mathbf{A} < \mathbf{B}$ have the same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_1$ be 4. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$, then the area of the quadrilateral $ABCD$ equals:

Step 1: Determine parameters for Ellipse E1

We are given that the eccentricity of ellipse $E_1$ is $e = \frac{1}{\sqrt{3}}$ and the distance between its foci is $2ae = 4$. We can use these to find the values of $a$ and $b$ for $E_1$. Recall that for an ellipse with $a>b$, $b^2 = a^2(1-e^2)$.

Step 2: Calculate 'a' and 'b' for E1

From $2ae = 4$ and $e = \frac{1}{\sqrt{3}}$, we find $a = \frac{4}{2e} = \frac{2}{1/\sqrt{3}} = 2\sqrt{3}$. Then, using $b^2 = a^2(1-e^2)$, we calculate $b^2 = (2\sqrt{3})^2(1 - (1/\sqrt{3})^2) = 12(1 - 1/3) = 12(2/3) = 8$. So, $b = \sqrt{8} = 2\sqrt{2}$.

Step 3: Determine parameters for Ellipse E2

For ellipse $E_2$, since $A < B$, the major axis is along the $y$ -axis. Its eccentricity is $e' = \frac{1}{\sqrt{3}}$, and $A^2 = B^2(1-e'^2)$. The length of the latus rectum for $E_1$ is $L_1 = \frac{2b^2}{a}$ and for $E_2$ is $L_2 = \frac{2A^2}{B}$. We are given that $L_1 L_2 = \frac{32}{\sqrt{3}}$.

Step 4: Calculate 'A' and 'B' for E2

Substitute the values of $a$ and $b^2$ for $E_1$ into the product of latus rectums equation: $\frac{2A^2}{B} \cdot \frac{2(8)}{2\sqrt{3}} = \frac{32}{\sqrt{3}}$. This simplifies to $\frac{16A^2}{B\sqrt{3}} = \frac{32}{\sqrt{3}}$, which gives $A^2 = 2B$. Also, for $E_2$, $A^2 = B^2(1-e'^2) = B^2(1 - (1/\sqrt{3})^2) = B^2(2/3)$. Equating the two expressions for $A^2$: $2B = B^2(2/3) \implies B = 3$. Then $A^2 = 2(3) = 6$, so $A = \sqrt{6}$.

Step 5: Find intersection points and area of quadrilateral ABCD

The equations of the ellipses are $E_1: \frac{x^2}{12} + \frac{y^2}{8} = 1$ and $E_2: \frac{x^2}{6} + \frac{y^2}{9} = 1$. To find the intersection points, we solve these equations simultaneously. Multiply the second equation by 2: $\frac{x^2}{3} + \frac{2y^2}{9} = 2$. Subtract this from the first equation: $(\frac{x^2}{12} - \frac{x^2}{3}) + (\frac{y^2}{8} - \frac{2y^2}{9}) = 1 - 2$. This gives $-\frac{3x^2}{12} + \frac{9y^2 - 16y^2}{72} = -1 \implies -\frac{x^2}{4} - \frac{7y^2}{72} = -1 \implies \frac{x^2}{4} + \frac{7y^2}{72} = 1$. This is incorrect. Let's solve for $x^2$ and $y^2$ directly. From $E_1$, $x^2 = 12(1 - y^2/8)$. From $E_2$, $x^2 = 6(1 - y^2/9)$. Equating them: $12(1 - y^2/8) = 6(1 - y^2/9) \implies 2(1 - y^2/8) = 1 - y^2/9 \implies 2 - y^2/4 = 1 - y^2/9 \implies 1 = y^2/4 - y^2/9 = \frac{9y^2 - 4y^2}{36} = \frac{5y^2}{36}$. So $y^2 = \frac{36}{5} \implies y = \pm \frac{6}{\sqrt{5}}$. Substitute $y^2$ back into $x^2 = 6(1 - y^2/9) = 6(1 - \frac{36}{5 \cdot 9}) = 6(1 - \frac{4}{5}) = 6(\frac{1}{5}) = \frac{6}{5}$. So $x = \pm \frac{\sqrt{6}}{\sqrt{5}}$. The intersection points are $(\pm \frac{\sqrt{6}}{\sqrt{5}}, \pm \frac{6}{\sqrt{5}})$. These points form a rectangle with vertices $(x_0, y_0), (-x_0, y_0), (-x_0, -y_0), (x_0, -y_0)$. The area of this rectangle is $2|x_0| \cdot 2|y_0| = 4|x_0 y_0|$. Area $= 4 \cdot \frac{\sqrt{6}}{\sqrt{5}} \cdot \frac{6}{\sqrt{5}} = 4 \cdot \frac{6\sqrt{6}}{5} = \frac{24\sqrt{6}}{5}$.

More from this chapter

• Find the equation of the circle passing through points (1,2), (3,4), and (5,2).
• Determine the equation of the circle with radius 4 and Centre (-2, 3).
• Find the equation of the hyperbola where foci are (0, +-6) and the length of the latus rectum is 36.
• Find the equation of the hyperbola where foci are (0, ± 12) and the length of the latus rectum is 36
• Find the equation of the circle with centre (-3,2) and radius 10.
• Find the equation of the circle with centre (-3,2) and radius 9.
• Find the equation of the circle with centre (-3,2) and radius 8.
• Find the equation of the circle with centre (-3,2) and radius 6.

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