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

Q: Why is the hyperbola formula $1 + \frac{b^2}{a^2}$ instead of $1 - \frac{b^2}{a^2}$?

Excellent question! The eccentricity measures how far the foci are from the center relative to the semi-major axis. In an ellipse, the foci are inside the vertices, so $c a$, leading to $e > 1$. The plus sign is mathematically necessary to ensure $e > 1$.

Q: Do I need to worry about which is $a^2$ and which is $b^2$ in the hyperbola?

In an ellipse, $a^2$ is always the larger denominator. In a hyperbola, $a^2$ is the denominator of the positive term (the transverse axis), regardless of which is larger. Always identify the positive term first.

Q: Why can't we just use energy conservation or some other physics principle?

While physics is beautiful, this is a problem of pure geometry. Conic sections are defined by their algebraic equations. We are mapping the 'shape' of one curve onto the 'shape' of another. Stick to the coordinate geometry definitions; they are the most direct path to the solution.

Question

If $e_1$ and $e_2$ are 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 if the point $(e_1, e_2)$ lies on ellipse $15x^2 + 3y^2 = k$. Then find value of $k$

Options:

(A) 14

(B) 15

(C) 16

(D) 17

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are not just solving an equation; we are exploring the fundamental DNA of conic sections. When you look at an ellipse or a hyperbola, you are looking at the paths traced by the laws of nature.

The eccentricity, denoted by $e$, is the parameter that defines the 'shape' of these curves. It tells us how much a conic section deviates from being a perfect circle.

### Phase 1: The Ellipse

Let us begin with the ellipse:

$$\frac{x^2}{18} + \frac{y^2}{4} = 1$$

Because the denominator under $x^2$ is larger than the one under $y^2$, we know this ellipse is stretched horizontally. The standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 18$ and $b^2 = 4$.

The eccentricity $e_1$ is our measure of its flatness. Using the formula $e_1 = \sqrt{1 - \frac{b^2}{a^2}}$, we substitute our values:

$$e_1 = \sqrt{1 - \frac{4}{18}} = \sqrt{1 - \frac{2}{9}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$$

This is the first coordinate of our point.

### Phase 2: The Hyperbola

Now, shift your focus to the hyperbola:

$$\frac{x^2}{9} - \frac{y^2}{4} = 1$$

Unlike the ellipse, which is a closed loop, the hyperbola is an open curve that races toward infinity. Its eccentricity $e_2$ must be greater than $1$.

We use the formula $e_2 = \sqrt{1 + \frac{b^2}{a^2}}$. Here, $a^2 = 9$ and $b^2 = 4$. Substituting these, we find:

$$e_2 = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$$

We have now found the second coordinate of our point.

### Phase 3: The Synthesis

We have arrived at the point $P(e_1, e_2) = (\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3})$. This point is a bridge between the two shapes we just analyzed.

The problem states that this point lies on a third curve, an ellipse defined by $15x^2 + 3y^2 = k$. If a point lies on a curve, it must satisfy the equation of that curve.

We substitute our coordinates into the equation:

$$15\left(\frac{\sqrt{7}}{3}\right)^2 + 3\left(\frac{\sqrt{13}}{3}\right)^2 = k$$

### Phase 4: The Final Calculation

Let us proceed with care. Squaring the terms, we get:

$$15\left(\frac{7}{9}\right) + 3\left(\frac{13}{9}\right) = k$$

This simplifies to:

$$\frac{105}{9} + \frac{39}{9} = k$$

Adding the numerators, we find $\frac{144}{9} = k$. Performing the final division, we arrive at the final result:

**$k = 16$**

It is elegant, isn't it? The complexity of the square roots vanishes, leaving us with a clean, integer result. This is the beauty of mathematics—when you follow the logic, the chaos resolves into order.

Select Answer:

Visualized Solution

Identify the First Ellipse

Formula for Ellipse Eccentricity $e_{1}$

Calculating $e_{1}$

Simplify $e_{1}$

Identify the Hyperbola

Formula for Hyperbola Eccentricity $e_{2}$

Calculating $e_{2}$

The Coordinate Point $(e_{1}, e_{2})$

The Third Ellipse Equation

Substituting the Values

Squaring Terms

Simplifying the Fractions

Final Calculation for $k$

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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