Let each of the two ellipses E1 : x²/a² + y²/b² = 1, (a b) and E2 : x²/A² + y²/B² = 1, (A

Q: Wait, why is the latus rectum formula $2A^2/B$ for the second ellipse? Shouldn't it always be $2b^2/a$?

That is a brilliant question! The formula $2b^2/a$ is just a placeholder for $2 \times \frac{(\text{semi-minor})^2}{\text{semi-major}}$. For $E_1$, the semi-major axis is $a$ (horizontal). For $E_2$, the semi-major axis is $B$ (vertical). Always remember: the latus rectum is always $2 \times \frac{\text{minor}^2}{\text{major}}$. Never memorize the variables; memorize the geometry!

Q: Why is the distance between foci $2Be$ for $E_2$?

Great observation. The distance between foci is always $2 \times \text{semi-major} \times e$. Since $E_2$ is vertical, its major axis is $2B$. Thus, the distance is $2Be$. If you used $2Ae$, you would be calculating the distance between foci on the wrong axis! Always identify the major axis first.

Q: Can I just use $e^2 = 1 - b^2/a^2$ for both?

You can, but you must be careful with the definition of $a$ and $b$. The formula is always $e^2 = 1 - \frac{(\text{semi-minor})^2}{(\text{semi-major})^2}$. If you blindly plug in $a$ and $b$ without checking which is the major axis, you will get a negative value under the square root or an incorrect eccentricity. Always check the orientation first!

Question

Let each of the two ellipses $E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$ and $E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, (A < B)$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $E_1$ and $E_2$ be $l_1$ and $l_2$, respectively, such that $2l_1^2 = 9l_2$. If the distance between the foci of $E_1$ is 8, then the distance between the foci of $E_2$ is

Options:

(A) $\frac{32}{5}$

(B) $\frac{8}{5}$

(C) $\frac{16}{5}$

(D) $\frac{96}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The ellipse is a shape of profound beauty—the path of planets, the cross-section of a cone, and a fundamental building block of our universe. When we look at two ellipses, $E_1$ and $E_2$, we are looking at two distinct geometric entities that share a common DNA: their eccentricity, $e = \frac{4}{5}$.

### Unlocking the Horizontal Ellipse

We begin with $E_1$, defined by the equation:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b $$

The problem states that the distance between the foci is $8$. In the language of geometry, this distance is $2ae$.

Given $2ae = 8$ and $e = \frac{4}{5}$, we find the semi-major axis $a$:

$$ 2a \left( \frac{4}{5} ight) = 8 \implies \frac{8a}{5} = 8 \implies a = 5 $$

Now, we determine the semi-minor axis $b$ using the fundamental relation $b^2 = a^2(1 - e^2)$:

$$ b^2 = 25 \left( 1 - \left( \frac{4}{5} ight)^2 ight) = 25 \left( 1 - \frac{16}{25} ight) = 25 \left( \frac{9}{25} ight) = 9 $$

The length of the latus rectum $l_1$ for $E_1$ is given by $\frac{2b^2}{a}$:

$$ l_1 = \frac{2(9)}{5} = \frac{18}{5} $$

### The Vertical Shift

Now, we turn our attention to $E_2$, defined by $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, where $A < B$. This is a vertical ellipse where the major axis lies along the $y$-axis.

For a vertical ellipse, the relation between the axes and eccentricity is $A^2 = B^2(1 - e^2)$. Substituting $e = \frac{4}{5}$:

$$ A^2 = B^2 \left( 1 - \frac{16}{25} ight) = \frac{9B^2}{25} $$

The latus rectum $l_2$ for this vertical ellipse is $\frac{2A^2}{B}$. Substituting our expression for $A^2$:

$$ l_2 = \frac{2}{B} \left( \frac{9B^2}{25} ight) = \frac{18B}{25} $$

### The Final Convergence

We are given the condition $2l_1^2 = 9l_2$. Substituting our derived expressions:

$$ 2 \left( \frac{18}{5} ight)^2 = 9 \left( \frac{18B}{25} ight) $$

Expanding the left side:

$$ 2 \left( \frac{324}{25} ight) = \frac{162B}{25} \implies \frac{648}{25} = \frac{162B}{25} $$

Canceling the denominators and solving for $B$:

$$ 648 = 162B \implies B = 4 $$

The distance between the foci of $E_2$ is $2Be$. Substituting $B = 4$ and $e = \frac{4}{5}$:

$$ 	ext{Distance} = 2(4) \left( \frac{4}{5} ight) = \frac{32}{5} $$

The final distance between the foci of $E_2$ is **$\frac{32}{5}$**.

Select Answer:

Visualized Solution

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

Foci Distance of $E_{1}$

Finding Semi-Major Axis $a$

Solving for $a$

Finding $b^{2}$ for $E_{1}$

Calculating $b^{2}$

Latus Rectum of $E_{1}$ ( $l_{1}$ )

Analyzing Vertical Ellipse $E_{2}$

Latus Rectum of $E_{2}$ ( $l_{2}$ )

Applying the Condition

Solving for $B$

Distance Between Foci of $E_{2}$

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

Analyzing the Setup

Unlocking the Horizontal Ellipse

The Vertical Shift

The Final Convergence

Similar Questions

Let the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Let the foci and length of the latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be $(\pm 5, 0)$ and $50$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x ^{2}}{b ^{2}} - \frac{y ^{2}}{a ^{2} b ^{2}} = 1$ equals

For some $θ \in (0, \frac{π}{2})$ , if the eccentricity of the hyperbola, $x^{2} - y^{2} sec^{2} θ = 10$ is $5$ times the eccentricity of the ellipse, $x^{2} sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is :

Select Answer:

Visualized Solution

Geometry 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​

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

Finding Semi-Major Axis .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; } a

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

Latus Rectum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​ ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l1​)

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

Latus Rectum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​ ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​)