Understanding the Problem

This problem asks us to find the value of a constant $k$ by connecting three different conic sections: an ellipse, a hyperbola, and another ellipse. We are given the equations of an ellipse and a hyperbola, and we need to calculate their respective eccentricities, $e_1$ and $e_2$. These eccentricities then form a point $(e_1, e_2)$, which is stated to lie on a third ellipse, $15x^2 + 3y^2 = k$. Our task is to use this information to determine the value of $k$. This requires a strong understanding of how to find the eccentricity of both ellipses and hyperbolas from their standard equations.

Key Concepts and Formulas

To solve this problem, we need to recall the standard forms and eccentricity formulas for ellipses and hyperbolas.

1. Standard Equation and Eccentricity of an Ellipse: The general equation of an ellipse centered at the origin is given by $\frac{x^2}{A} + \frac{y^2}{B} = 1$.

   • Identifying $a^2$ and $b^2$: For an ellipse, $a^2$ always represents the square of the semi-major axis, and $b^2$ represents the square of the semi-minor axis. This means $a^2$ is always the larger of the two denominators ($A$ and $B$), and $b^2$ is the smaller one.
     • If the major axis is along the x-axis, then $a^2 = A$ and $b^2 = B$ (where $A > B$).
     • If the major axis is along the y-axis, then $a^2 = B$ and $b^2 = A$ (where $B > A$).
   • Eccentricity Formula: The eccentricity $e$ of an ellipse is given by the formula: $e = \sqrt{1 - \frac{b^2}{a^2}}$ Why this formula? Eccentricity measures how "stretched" an ellipse is. For an ellipse, $0 \le e < 1$. The term $\frac{b^2}{a^2}$ is always a positive fraction less than 1 (since $b^2 < a^2$), ensuring that $1 - \frac{b^2}{a^2}$ is between 0 and 1, leading to $e < 1$.

2. Standard Equation and Eccentricity of a Hyperbola: The general equation of a hyperbola centered at the origin can be one of two forms:

   • Transverse axis along x-axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
   • Transverse axis along y-axis: $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
   • Identifying $a^2$ and $b^2$: For a hyperbola, $a^2$ is always the denominator of the positive term, and $b^2$ is always the denominator of the negative term. Unlike an ellipse, $a^2$ and $b^2$ are not determined by their relative magnitudes.
   • Eccentricity Formula: The eccentricity $e$ of a hyperbola is given by the formula:
     • For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $e = \sqrt{1 + \frac{b^2}{a^2}}$
     • For $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$: $e = \sqrt{1 + \frac{a^2}{b^2}}$ Why this formula? For a hyperbola, $e > 1$. The term $\frac{b^2}{a^2}$ (or $\frac{a^2}{b^2}$) is always positive, so $1 + \text{positive term}$ is always greater than 1, ensuring $e > 1$.

3. Point on a Curve: If a point $(x_0, y_0)$ lies on a curve defined by an algebraic equation, substituting $x_0$ for $x$ and $y_0$ for $y$ into the equation will satisfy the equation. This is a fundamental principle in coordinate geometry.

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Step-by-Step Solution

Our strategy involves three main steps:

1. Calculate $e_1$ for the given ellipse.
2. Calculate $e_2$ for the given hyperbola.
3. Substitute $(e_1, e_2)$ into the equation of the third ellipse to find $k$.

Step 1: Calculate the eccentricity of the ellipse ($e_1$).

The equation of the first curve is given as: $\frac{x^2}{18} + \frac{y^2}{4} = 1$

• Identify $a^2$ and $b^2$: We compare this to the standard ellipse form $\frac{x^2}{A} + \frac{y^2}{B} = 1$. Here, $A=18$ and $B=4$. For an ellipse, $a^2$ is the larger denominator and $b^2$ is the smaller denominator. Since $18 > 4$, we have: $a^2 = 18$ $b^2 = 4$

  • Why? This tells us that the major axis of this ellipse is along the x-axis, as the larger denominator is under the $x^2$ term.

• Apply the eccentricity formula: The formula for the eccentricity of an ellipse is $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substitute the values of $a^2$ and $b^2$: $e_1 = \sqrt{1 - \frac{4}{18}}$

• Simplify the expression: First, simplify the fraction $\frac{4}{18}$ by dividing both numerator and denominator by 2: $\frac{4}{18} = \frac{2}{9}$ Now, substitute this back into the eccentricity formula: $e_1 = \sqrt{1 - \frac{2}{9}}$ To combine the terms under the square root, find a common denominator: $e_1 = \sqrt{\frac{9}{9} - \frac{2}{9}}$ $e_1 = \sqrt{\frac{9 - 2}{9}}$ $e_1 = \sqrt{\frac{7}{9}}$ Finally, take the square root of the numerator and the denominator separately: $e_1 = \frac{\sqrt{7}}{\sqrt{9}}$ $e_1 = \frac{\sqrt{7}}{3}$ So, the eccentricity of the ellipse is $e_1 = \frac{\sqrt{7}}{3}$. (Note: $\frac{\sqrt{7}}{3} \approx \frac{2.64}{3} < 1$, which is consistent for an ellipse).

Step 2: Calculate the eccentricity of the hyperbola ($e_2$).

The equation of the second curve is given as: $\frac{x^2}{9} - \frac{y^2}{4} = 1$

• Identify $a^2$ and $b^2$: We compare this to the standard hyperbola form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. For a hyperbola, $a^2$ is the denominator of the positive term, and $b^2$ is the denominator of the negative term. From the given equation: $a^2 = 9$ $b^2 = 4$

  • Why? In a hyperbola, the parameters $a$ and $b$ are defined by their position relative to the positive/negative terms, not by their magnitude. The $x^2$ term is positive, so its denominator is $a^2$. The $y^2$ term is negative, so its denominator is $b^2$.

• Apply the eccentricity formula: The formula for the eccentricity of a hyperbola of this form is $e = \sqrt{1 + \frac{b^2}{a^2}}$. Substitute the values of $a^2$ and $b^2$: $e_2 = \sqrt{1 + \frac{4}{9}}$

• Simplify the expression: To combine the terms under the square root, find a common denominator: $e_2 = \sqrt{\frac{9}{9} + \frac{4}{9}}$ $e_2 = \sqrt{\frac{9 + 4}{9}}$ $e_2 = \sqrt{\frac{13}{9}}$ Finally, take the square root of the numerator and the denominator separately: $e_2 = \frac{\sqrt{13}}{\sqrt{9}}$ $e_2 = \frac{\sqrt{13}}{3}$ So, the eccentricity of the hyperbola is $e_2 = \frac{\sqrt{13}}{3}$. (Note: $\frac{\sqrt{13}}{3} \approx \frac{3.6}{3} > 1$, which is consistent for a hyperbola).

Step 3: Substitute the eccentricities into the equation of the third ellipse to find $k$.

We are given that the point $(e_1, e_2)$ lies on the ellipse $15x^2 + 3y^2 = k$.

• Why? If a point lies on a curve, its coordinates must satisfy the equation of that curve. Therefore, we can substitute $x=e_1$ and $y=e_2$ into the equation.

• Substitute the values: $15(e_1)^2 + 3(e_2)^2 = k$ Substitute the values we found for $e_1$ and $e_2$: $15\left(\frac{\sqrt{7}}{3}\right)^2 + 3\left(\frac{\sqrt{13}}{3}\right)^2 = k$

• Square the terms: When squaring a fraction, we square both the numerator and the denominator: $\left(\frac{\sqrt{7}}{3}\right)^2 = \frac{(\sqrt{7})^2}{3^2} = \frac{7}{9}$ $\left(\frac{\sqrt{13}}{3}\right)^2 = \frac{(\sqrt{13})^2}{3^2} = \frac{13}{9}$ Substitute these squared values back into the equation: $15\left(\frac{7}{9}\right) + 3\left(\frac{13}{9}\right) = k$

• Perform the multiplication: Multiply the whole numbers by the numerators of the fractions: $\frac{15 \times 7}{9} + \frac{3 \times 13}{9} = k$ $\frac{105}{9} + \frac{39}{9} = k$

• Add the fractions: Since the fractions have a common denominator, we can simply add their numerators: $\frac{105 + 39}{9} = k$ $\frac{144}{9} = k$

• Calculate the final value of $k$: Divide 144 by 9: $k = 16$

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Tips for Success and Common Pitfalls

• Distinguishing $a^2$ and $b^2$ for Ellipse vs. Hyperbola: This is the most critical point.
  • Ellipse: $a^2$ is the larger denominator, $b^2$ is the smaller.
  • Hyperbola: $a^2$ is the denominator of the positive term, $b^2$ is the denominator of the negative term, regardless of their relative magnitudes.
• Eccentricity Formula Signs: Remember the key difference:
  • Ellipse: $e = \sqrt{1 - \frac{\text{smaller}}{\text{larger}}}$ (minus sign, $e<1$)
  • Hyperbola: $e = \sqrt{1 + \frac{\text{denominator of negative term}}{\text{denominator of positive term}}}$ (plus sign, $e>1$)

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