Solution: Finding the Sum of Squares of Semi-Axes for an Ellipse

This problem asks us to determine the sum of the squares of the semi-major and semi-minor axes ($a^2 + b^2$) for a given ellipse. We are provided with its eccentricity and a point through which it passes. To solve this, we will leverage the fundamental properties and equations of an ellipse.

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1. Understanding the Key Properties of an Ellipse

Before diving into the calculations, let's recall the essential definitions and formulas for an ellipse centered at the origin, as described in the problem:

• Standard Equation of an Ellipse (Major Axis along X-axis): The general equation for an ellipse centered at the origin, with its major axis lying along the x-axis (meaning $a > b$), is given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Here, $a$ represents the length of the semi-major axis (half the length of the longest diameter), and $b$ represents the length of the semi-minor axis (half the length of the shortest diameter). The condition $a > b$ is crucial as it dictates the orientation of the major axis and the specific eccentricity formula to use.

• Eccentricity of an Ellipse ($e$): Eccentricity is a measure of how "stretched out" an ellipse is. For an ellipse where the major axis is along the x-axis ($a > b$), the relationship between $a$, $b$, and $e$ is: $b^2 = a^2(1 - e^2)$ Alternatively, this can be written as $e^2 = 1 - \frac{b^2}{a^2}$. The value of $e$ for an ellipse always lies between 0 and 1 ($0 < e < 1$).

Our strategy will be to use the given eccentricity to establish a relationship between $a^2$ and $b^2$. Then, we will use the fact that the ellipse passes through a specific point to form a second equation involving $a^2$ and $b^2$. Finally, we will solve these two equations simultaneously to find the values of $a^2$ and $b^2$, and then calculate their sum.

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2. Step 1: Using the Eccentricity to Relate $a^2$ and $b^2$

We are given that the eccentricity of the ellipse is $e = \frac{1}{4}$. Since the problem specifies $a > b$, we use the standard eccentricity formula for an ellipse with its major axis along the x-axis.

• Action: Substitute the given value of $e$ into the eccentricity formula: $b^2 = a^2(1 - e^2)$

• Explanation: We substitute $e = \frac{1}{4}$ into the formula to establish a direct algebraic relationship between $a^2$ and $b^2$. This is our first equation.

• Calculation: $b^2 = a^2 \left( 1 - \left(\frac{1}{4}\right)^2 \right)$ $b^2 = a^2 \left( 1 - \frac{1}{16} \right)$ $b^2 = a^2 \left( \frac{16 - 1}{16} \right)$ $b^2 = a^2 \left( \frac{15}{16} \right)$

• Rearrangement for Substitution: To make it easier for substitution later, let's express $a^2$ in terms of $b^2$: $16b^2 = 15a^2$ $a^2 = \frac{16b^2}{15} \quad \text{(Equation 1)}$ This equation is crucial as it links our two unknown parameters.

• Tip: Always verify that you are using the correct eccentricity formula based on whether $a > b$ (major axis along x-axis) or $b > a$ (major axis along y-axis). The problem explicitly states $a > b$, confirming our choice.

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3. Step 2: Forming an Equation using the Given Point on the Ellipse

The problem states that the ellipse passes through the point $\left( -4\sqrt{\frac{2}{5}}, 3 \right)$. A fundamental property of any curve is that if a point lies on it, its coordinates must satisfy the curve's equation.

• Action: Substitute the $x$-coordinate $x = -4\sqrt{\frac{2}{5}}$ and the $y$-coordinate $y = 3$ into the standard equation of the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

• Explanation: By substituting the coordinates, we are creating a second equation involving $a^2$ and $b^2$, which will allow us to solve for their specific values.

• Calculation of Squared Terms: First, let's calculate the squares of the coordinates: $x^2 = \left( -4\sqrt{\frac{2}{5}} \right)^2 = (-4)^2 \times \left(\sqrt{\frac{2}{5}}\right)^2 = 16 \times \frac{2}{5} = \frac{32}{5}$ $y^2 = (3)^2 = 9$

• Substitution into Ellipse Equation: Now substitute these squared values into the ellipse's equation: $\frac{\frac{32}{5}}{a^2} + \frac{9}{b^2} = 1$

• Simplification: We can simplify the first term by moving the denominator of the numerator ($5$) to the main denominator: $\frac{32}{5a^2} + \frac{9}{b^2} = 1 \quad \text{(Equation 2)}$ This is our second algebraic relationship between $a^2$ and $b^2$.

• Common Mistake: Be meticulous with arithmetic, especially when squaring terms involving fractions and square roots. Errors here are frequent and can derail the entire solution.

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4. Step 3: Solving the System of Equations for $a^2$ and $b^2$

We now have a system of two algebraic equations with two unknowns ($a^2$ and $b^2$):

1. $a^2 = \frac{16b^2}{15}$
2. $\frac{32}{5a^2} + \frac{9}{b^2} = 1$

• Action: Substitute Equation (1) into Equation (2) to eliminate $a^2$ and solve for $b^2$.

• Explanation: This substitution method is efficient as Equation (1) already expresses $a^2$ in terms of $b^2$, simplifying the process of reducing the system to a single variable.

• Substitution and Simplification: Substitute $a^2 = \frac{16b^2}{15}$ into Equation (2): $\frac{32}{5 \left( \frac{16b^2}{15} \right)} + \frac{9}{b^2} = 1$ Simplify the denominator of the first term: $5 \times \frac{16b^2}{15} = \frac{80b^2}{15}$. We can simplify this fraction by dividing both numerator and denominator by 5: $\frac{16b^2}{3}$. So the equation becomes: $\frac{32}{\frac{16b^2}{3}} + \frac{9}{b^2} = 1$ To simplify the complex fraction $\frac{32}{\frac{16b^2}{3}}$, multiply 32 by the reciprocal of the denominator: $32 \times \frac{3}{16b^2} + \frac{9}{b^2} = 1$ $\frac{32 \times 3}{16b^2} + \frac{9}{b^2} = 1$ We can simplify $\frac{32}{16}$ to $2$: $\frac{2 \times 3}{b^2} + \frac{9}{b^2} = 1$ $\frac{6}{b^2} + \frac{9}{b^2} = 1$

• Combine terms and Solve for $b^2$: Since both terms have the same denominator $b^2$, we can combine their numerators: $\frac{6 + 9}{b^2} = 1$ $\frac{15}{b^2} = 1$ $b^2 = 15$

• Find $a^2$: Now that we have the value of $b^2$, substitute $b^2 = 15$ back into Equation (1) to find $a^2$: $a^2 = \frac{16b^2}{15}$ $a^2 = \frac{16(15)}{15}$ $a^2 = 16$

• Verification: We found $a^2 = 16$ and $b^2 = 15$. This means $a = \sqrt{16} = 4$ and $b = \sqrt{15}$. Since $4 > \sqrt{15}$ (as $16 > 15$), the condition $a > b$ given in the question is satisfied, confirming the validity of our calculations and the formulas used.

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5. Step 4: Calculate the Required Sum $a^2 + b^2$

The question asks for the value of $a^2 + b^2$.

• Action: Add the values of $a^2$ and $b^2$ that we have just determined.

• Calculation: $a^2 + b^2 = 16 + 15$ $a^2 + b^2 = 31$

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6. Conclusion and Final Answer

The calculated value of $a^2 + b^2$ is 31.

Comparing this with the given options: (A) 29 (B) 31 (C) 32 (D) 34

Our result matches option (B).

Summary and Key Takeaway: This problem is a classic example of how to solve coordinate geometry questions involving conic sections. The systematic approach involves:

1. Identifying Key Formulas: Correctly recalling the standard equation and eccentricity relation for the specified conic section (ellipse with $a>b$).
2. Formulating Equations: Using all provided information (eccentricity value, points on the curve) to construct a system of algebraic equations with the unknown parameters ($a^2, b^2$).
3. Solving Systematically: Applying algebraic techniques (like substitution) to solve the system of equations.
4. Verifying Conditions: Always checking if the derived parameters satisfy any initial conditions or constraints (e.g., $a>b$). This structured method ensures accuracy and clarity in reaching the solution.

The final answer is $\boxed{\text{31}}$.