If e1 and e2 are the eccentricities of the ellipse, x²/18 + y²/4 = 1 and the hyperbola, x²/9 - y²/4

Q: Wait, why is the hyperbola eccentricity formula different? I keep mixing them up!

It is a common hurdle! Think of the ellipse as a 'contained' shape where $e 1$, so we add the ratio. The plus sign is the key to its infinite reach.

Q: Do I need to worry about the sign of the coordinates $(e_1, e_2)$?

Excellent question! Eccentricity is defined as a ratio of lengths, so it is always positive. You are working in the first quadrant, which simplifies things immensely.

Q: Is there a shortcut for the calculation?

While there is no 'magic' shortcut, keeping the denominators as $9$ throughout your calculation is a powerful strategy. It prevents premature rounding and makes the final addition trivial.

Question

If $e_1$ and $e_2$ are the eccentricities of the ellipse, $\frac{x^2}{18} + \frac{y^2}{4} = 1$ and the hyperbola, $\frac{x^2}{9} - \frac{y^2}{4} = 1$ respectively and $(e_1, e_2)$ is a point on the ellipse, $15x^2 + 3y^2 = k$. Then $k$ is equal to:

Options:

(A) $14$

(B) $15$

(C) $17$

(D) $16$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe! Today, we are going to unravel a beautiful problem that bridges the gap between the elegant curves of an ellipse and the sweeping, infinite arms of a hyperbola.

We are not just solving for a constant $k$; we are uncovering the hidden relationship between the eccentricities of these two fundamental conic sections. Let us begin by visualizing our players.

We have an ellipse, 
$$ \frac{x^2}{18} + \frac{y^2}{4} = 1 $$ 
and a hyperbola, 
$$ \frac{x^2}{9} - \frac{y^2}{4} = 1 $$

Our goal is to find the value of $k$ for the curve $15x^2 + 3y^2 = k$ that passes through the point $(e_1, e_2)$, where $e_1$ and $e_2$ are the eccentricities of our first two curves.

### Phase 1: The Ellipse and its Eccentricity

First, let us look at our ellipse: 
$$ \frac{x^2}{18} + \frac{y^2}{4} = 1 $$

By comparing this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we immediately see that $a^2 = 18$ and $b^2 = 4$. The eccentricity $e_1$ of an ellipse is defined by the formula:
$$ e_1^2 = 1 - \frac{b^2}{a^2} $$

Substituting our values, we get $e_1^2 = 1 - \frac{4}{18}$. Simplifying the fraction $\frac{4}{18}$ gives us $\frac{2}{9}$.

Thus, $e_1^2 = 1 - \frac{2}{9} = \frac{7}{9}$, which means $e_1 = \frac{\sqrt{7}}{3}$. This is our first coordinate.

### Phase 2: The Hyperbola and its Eccentricity

Now, let us turn our attention to the hyperbola: 
$$ \frac{x^2}{9} - \frac{y^2}{4} = 1 $$

Here, $a^2 = 9$ and $b^2 = 4$. The eccentricity $e_2$ of a hyperbola is defined by:
$$ e_2^2 = 1 + \frac{b^2}{a^2} $$

Substituting our values, we get $e_2^2 = 1 + \frac{4}{9} = \frac{13}{9}$. Therefore, $e_2 = \frac{\sqrt{13}}{3}$. We have successfully found our second coordinate.

### Phase 3: The Final Convergence

We are now at the heart of the problem. We have a point $P = (e_1, e_2) = (\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3})$ that lies on the curve $15x^2 + 3y^2 = k$.

Because this point lies on the curve, it must satisfy the equation. Let us substitute our values:
$$ 15\left(\frac{\sqrt{7}}{3}\right)^2 + 3\left(\frac{\sqrt{13}}{3}\right)^2 = k $$

Squaring these terms is straightforward: $\left(\frac{\sqrt{7}}{3}\right)^2 = \frac{7}{9}$ and $\left(\frac{\sqrt{13}}{3}\right)^2 = \frac{13}{9}$. Now, our equation becomes:
$$ 15\left(\frac{7}{9}\right) + 3\left(\frac{13}{9}\right) = k $$

Multiplying these out, we get:
$$ \frac{105}{9} + \frac{39}{9} = k $$

Adding the numerators, we find $\frac{144}{9} = k$. Dividing $144$ by $9$ yields exactly $16$.

And there it is! The constant **$k = 16$**. It is a beautiful, clean result that emerges from the interplay of these conic sections.

Select Answer:

Visualized Solution

The Three Curves

Parameters of the First Ellipse

Eccentricity Formula for Ellipse

Calculate $e_{1}$

Parameters of the Hyperbola

Eccentricity Formula for Hyperbola

Calculate $e_{2}$

Locating the Point $(e_{1}, e_{2})$

Substituting the Point

Squaring the Terms

Simplifying the Expression

Final Calculation for $k$

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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