Let E: x²/a² + y²/b² = 1 ab and H: x²/A² - y²/B² = 1 Let the distance between the foci of E and the

Q: Why do we assume the distance from the center to the focus is $ae$ for the ellipse and $Ae'$ for the hyperbola?

That is the fundamental definition! In an ellipse, the foci are located at $(\pm ae, 0)$. In a hyperbola, they are located at $(\pm Ae', 0)$. It is the geometric DNA of these curves. When the problem states the distance between foci is $2\sqrt{3}$, it is telling you that $2ae = 2\sqrt{3}$ and $2Ae' = 2\sqrt{3}$.

Q: Is it possible for the latus rectum lengths to be equal? It feels like a coincidence.

In mathematics, especially in JEE problems, 'coincidence' is often a sign of elegant symmetry. Here, the specific constraints on the focal distance and eccentricity ratio force the geometry to align such that the latus rectums are equal. It is not a coincidence; it is the beautiful result of the constraints provided.

Q: Why can't we just use the formula for the latus rectum directly?

Because you don't know $a, b, A,$ or $B$ yet! You have the relationships between them, but not the values. You must solve for the parameters first, then calculate the latus rectum lengths. Think of it as peeling an onion—you must remove the outer layers of unknowns before you reach the core values.

Question

Let E: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $a>b$ and H: $\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1$ Let the distance between the foci of E and the foci of H be $2\sqrt{3}$. If $a-A=2$ and the ratio of the eccentricities of E and H is $\frac{1}{3}$ then the sum of the lengths of their latus rectums is equal to:

Options:

(A) 10

(B) 9

(C) 8

(D) 7

Sigma Percentile · Accepted Answer

### Analyzing the Shared Geometry

The problem states that the ellipse $E$ and the hyperbola $H$ share the same focal points. The distance between the foci for both curves is given as $2\sqrt{3}$.

For an ellipse with semi-major axis $a$ and eccentricity $e$, the distance between foci is $2ae$. For a hyperbola with semi-major axis $A$ and eccentricity $e'$, the distance between foci is $2Ae'$.

Setting these equal to the given value, we obtain:
$$ae = \sqrt{3}$$
$$Ae' = \sqrt{3}$$

### The Eccentricity Bridge

We are given that the ratio of the eccentricity of the ellipse to the hyperbola is $\frac{1}{3}$. This implies the relationship:
$$e' = 3e$$

Substituting this into our hyperbola focal distance equation, we get $A(3e) = \sqrt{3}$. Comparing this to the ellipse equation $ae = \sqrt{3}$, we can equate the two expressions:
$$3Ae = ae$$

By canceling the common term $e$, we find the relationship between the semi-major axes:
$$a = 3A$$

### The Algebraic Unlocking

We are provided with the constraint $a - A = 2$. Substituting $a = 3A$ into this equation yields:
$$3A - A = 2$$
$$2A = 2 \Rightarrow A = 1$$

Consequently, we find the semi-major axis of the ellipse:
$$a = 3(1) = 3$$

With these values, we verify the eccentricities:
$$e = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$
$$e' = \frac{\sqrt{3}}{1} = \sqrt{3}$$

### Calculating the Latus Rectum

The length of the latus rectum for an ellipse is given by $L_E = \frac{2b^2}{a}$. Using the identity $b^2 = a^2(1 - e^2)$, we calculate:
$$b^2 = 9 \left( 1 - \frac{1}{3} \right) = 9 \left( \frac{2}{3} \right) = 6$$
$$L_E = \frac{2(6)}{3} = 4$$

The length of the latus rectum for a hyperbola is given by $L_H = \frac{2B^2}{A}$. Using the identity $B^2 = A^2(e'^2 - 1)$, we calculate:
$$B^2 = 1(3 - 1) = 2$$
$$L_H = \frac{2(2)}{1} = 4$$

### Final Calculation

The sum of the lengths of the latus rectums of the ellipse and the hyperbola is:
$$L_E + L_H = 4 + 4 = 8$$

The final result is **8**.

Select Answer:

Visualized Solution

Visualizing the Curves

Distance Between Foci

Simplifying Foci Relations

Ratio of Eccentricities

Relating Semi-Major Axes

Solving for $a$ and $A$

Calculating Eccentricities

Finding $b^{2}$ for Ellipse

Finding $B^{2}$ for Hyperbola

Latus Rectum of Ellipse

Latus Rectum of Hyperbola

Final Sum of Lengths

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Shared Geometry

The Eccentricity Bridge

The Algebraic Unlocking

Calculating the Latus Rectum

Final Calculation

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