Let the ellipse E₁: x²/a² + y²/b² = 1 ab and E₂: x²/A² + y²/B² = 1, A

Q: Wait, why is the latus rectum formula different for $E_{2}$?

Great observation! The formula for the latus rectum depends on the major axis. For a horizontal ellipse, the major axis is along the x-axis, so the latus rectum is $\frac{2b^{2}}{a}$. For a vertical ellipse, the major axis is along the y-axis, so the latus rectum is $\frac{2A^{2}}{B}$. Always check the orientation ($a > b$ vs $A < B$) before applying the formula!

Q: How do we know the intersection points form a rectangle?

Because both ellipses are centered at the origin and are symmetric with respect to both the x-axis and the y-axis. If $(x, y)$ is an intersection point, then $(-x, y)$, $(x, -y)$, and $(-x, -y)$ must also be intersection points. These four points form a rectangle with sides parallel to the coordinate axes.

Q: Can we use polar coordinates to solve this?

While you technically could, it would be significantly more complex due to the different orientations of the ellipses. Algebraic substitution is much more direct and less prone to calculation errors in this specific case.

Question

Let the ellipse $E_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $a>b$ and $E_{2}:\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$, $A<B$ have 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 :

Options:

(A) $\frac{12\sqrt{6}}{5}$

(B) $6\sqrt{6}$

(C) $\frac{18\sqrt{6}}{5}$

(D) $\frac{24\sqrt{6}}{5}$

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Unlocking the Ellipse

Imagine you are standing on a vast, flat plane, and before you lie two ellipses, $E_{1}$ and $E_{2}$. They are not just random shapes; they are governed by the same eccentricity, $e = \frac{1}{\sqrt{3}}$, yet they possess different orientations.

$E_{1}$ is a horizontal ellipse, stretching its arms along the x-axis, while $E_{2}$ is a vertical one, reaching towards the y-axis. Our mission is to find the area of the quadrilateral formed by their intersection.

### Phase 1: Decoding $E_{1}$

We begin with $E_{1}: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$. We are told that the distance between its foci is $4$. For a horizontal ellipse, this distance is defined by $2ae$.

With $e = \frac{1}{\sqrt{3}}$, we set up our first equation:
$$2a \left( \frac{1}{\sqrt{3}} ight) = 4$$

Solving this, we find $a = 2\sqrt{3}$, which means $a^{2} = 12$. Now, we need $b^{2}$. Using the fundamental relation $b^{2} = a^{2}(1 - e^{2})$, we calculate:
$$b^{2} = 12 \left( 1 - \frac{1}{3} ight) = 12 \left( \frac{2}{3} ight) = 8$$

With $a^{2} = 12$ and $b^{2} = 8$, the equation for $E_{1}$ is locked in:
$$\frac{x^{2}}{12} + \frac{y^{2}}{8} = 1 \quad \Rightarrow \quad 2x^{2} + 3y^{2} = 24$$

### Phase 2: The Bridge of Latus Recta

Next, we calculate the latus rectum of $E_{1}$, denoted as $L_{1} = \frac{2b^{2}}{a}$. Substituting our values:
$$L_{1} = \frac{2(8)}{2\sqrt{3}} = \frac{8}{\sqrt{3}}$$

The problem gives us a beautiful piece of information: the product of the latus rectums is $L_{1} \cdot L_{2} = \frac{32}{\sqrt{3}}$. By substituting $L_{1}$, we find:
$$L_{2} = \frac{32}{\sqrt{3}} \cdot \frac{\sqrt{3}}{8} = 4$$

### Phase 3: Decoding $E_{2}$

$E_{2}$ is a vertical ellipse ($A < B$), so its latus rectum is $L_{2} = \frac{2A^{2}}{B} = 4$, which simplifies to $A^{2} = 2B$. Its eccentricity relation is:
$$A^{2} = B^{2}(1 - e^{2}) = B^{2} \left( \frac{2}{3} ight)$$

Equating our two expressions for $A^{2}$, we get $2B = \frac{2}{3}B^{2}$. Since $B 
eq 0$, we find $B = 3$, so $B^{2} = 9$.

Consequently, $A^{2} = 2(3) = 6$. The equation for $E_{2}$ is:
$$\frac{x^{2}}{6} + \frac{y^{2}}{9} = 1 \quad \Rightarrow \quad 3x^{2} + 2y^{2} = 18$$

### Phase 4: The Intersection

We now have our two equations: $2x^{2} + 3y^{2} = 24$ and $3x^{2} + 2y^{2} = 18$. To find the intersection points, we solve this system.

Adding them gives $5x^{2} + 5y^{2} = 42$, so $x^{2} + y^{2} = \frac{42}{5}$. By eliminating $x^{2}$, we find:
$$y^{2} = \frac{36}{5} \quad 	ext{and} \quad x^{2} = \frac{6}{5}$$

The intersection points are $(\pm \sqrt{\frac{6}{5}}, \pm \sqrt{\frac{36}{5}})$.

### Conclusion: The Area

These four points form a rectangle. The width is $2x = 2\sqrt{\frac{6}{5}}$ and the height is $2y = 2\sqrt{\frac{36}{5}}$.

The area is:
$$Area = (2x)(2y) = 4xy = 4 \sqrt{\frac{6}{5}} \cdot \frac{6}{\sqrt{5}} = 4 \cdot \frac{6\sqrt{6}}{5} = \mathbf{\frac{24\sqrt{6}}{5}}$$

We have navigated the geometry, solved the algebra, and arrived at the elegant solution. You have mastered the ellipse!

Select Answer:

Visualized Solution

Visualize Ellipse $E_{1}$

Distance Between Foci

Setting up the Equation for $a$

Solving for $a^{2}$

Eccentricity Relation for $E_{1}$

Calculating $b^{2}$ and $L_{1}$

Introduce Ellipse $E_{2}$

Finding Latus Rectum $L_{2}$

Solving for $L_{2}$

Formulas for Vertical Ellipse $E_{2}$

Setting up Equations for $A$ and $B$

Solving for $A^{2}$ and $B^{2}$

Intersection of $E_{1}$ and $E_{2}$

Solving for Intersection Points

Area of Quadrilateral ABCD

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Select Answer:

Visualized Solution

Visualize Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​

Distance Between Foci

Setting up the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Eccentricity Relation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Introduce Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Finding Latus Rectum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Formulas for Vertical Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Setting up Equations for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2

Intersection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​