Key Concepts and Formulas

An ellipse is a conic section defined as the locus of points where the sum of the distances from two fixed points (called foci) is constant. For an ellipse centered at the origin $(0,0)$ with its axes aligned along the coordinate axes, its standard equation is given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Here, $a$ and $b$ represent the lengths of the semi-axes.

• If $a > b$, the major axis (the longer one) lies along the x-axis, and $a$ is the length of the semi-major axis. The minor axis (the shorter one) lies along the y-axis, and $b$ is the length of the semi-minor axis.
• If $b > a$, the major axis lies along the y-axis, and $b$ is the length of the semi-major axis. The minor axis lies along the x-axis, and $a$ is the length of the semi-minor axis.

The eccentricity, denoted by $e$, is a fundamental characteristic of an ellipse that quantifies its "roundness" or "elongation". For an ellipse, $0 < e < 1$. It is related to $a$ and $b$ by the following formulas:

• If $a > b$ (major axis along x-axis): $b^2 = a^2(1 - e^2)$
• If $b > a$ (major axis along y-axis): $a^2 = b^2(1 - e^2)$ A more general way to remember this is $e^2 = 1 - \frac{(\text{semi-minor axis})^2}{(\text{semi-major axis})^2}$.

Problem Analysis

We are tasked with finding the eccentricity of an ellipse. We are provided with the following crucial pieces of information:

1. Center at the origin: This immediately tells us the ellipse is of the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
2. Axes along the coordinate axes: This confirms that there are no rotation or translation terms in the equation, reinforcing the use of the standard form.
3. Passes through points $(4, -1)$ and $(-2, 2)$: These two points are key. Since they lie on the ellipse, their coordinates must satisfy the ellipse's equation. This will allow us to form a system of equations to determine the unknown parameters $a^2$ and $b^2$.

Our strategy will be to use the given points to find the values of $a^2$ and $b^2$. Once we know these values (or a relationship between them), we can determine which axis is the major axis and then apply the appropriate eccentricity formula.

Step-by-Step Solution

1. Formulate the General Equation of the Ellipse

Given that the ellipse is centered at the origin and its axes are along the coordinate axes, we can directly write its standard equation. This is our starting point because it encapsulates all the general properties mentioned in the problem statement. $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \text{(Equation A)}$ Here, $a^2$ and $b^2$ are constants that define the specific dimensions of our ellipse, and our primary goal is to find them.

2. Utilize the First Given Point (4, -1)

Since the ellipse passes through the point $(4, -1)$, this means that if we substitute $x=4$ and $y=-1$ into Equation A, the equation must hold true. We perform this substitution to generate our first equation relating $a^2$ and $b^2$. Substituting $x=4$ and $y=-1$: $\frac{(4)^2}{a^2} + \frac{(-1)^2}{b^2} = 1$ $\frac{16}{a^2} + \frac{1}{b^2} = 1$ To make this equation easier to work with, we clear the denominators by multiplying the entire equation by the least common multiple of $a^2$ and $b^2$, which is $a^2b^2$: $16b^2 + a^2 = a^2b^2 \quad \text{(Equation 1)}$ This equation now provides a specific relationship between $a^2$ and $b^2$ that must be satisfied for the ellipse to pass through $(4, -1)$.

3. Utilize the Second Given Point (-2, 2)

Similarly, the ellipse also passes through the point $(-2, 2)$. Just as with the first point, we substitute these coordinates into Equation A. This step is crucial for creating a second, independent equation, which will allow us to solve for both $a^2$ and $b^2$. Substituting $x=-2$ and $y=2$: $\frac{(-2)^2}{a^2} + \frac{(2)^2}{b^2} = 1$ $\frac{4}{a^2} + \frac{4}{b^2} = 1$ Again, we clear the denominators by multiplying by $a^2b^2$ to simplify the expression: $4b^2 + 4a^2 = a^2b^2 \quad \text{(Equation 2)}$ Now we have a system of two equations (Equation 1 and Equation 2) with two unknowns, $a^2$ and $b^2$.

4. Solve the System of Equations for $a^2$ and $b^2$

We have the following system: From Equation 1: $16b^2 + a^2 = a^2b^2$ From Equation 2: $4b^2 + 4a^2 = a^2b^2$

Notice that both Equation 1 and Equation 2 are equal to the term $a^2b^2$. This provides a straightforward way to solve the system: we can equate their left-hand sides. This eliminates $a^2b^2$ and leaves us with an equation solely in terms of $a^2$ and $b^2$. $16b^2 + a^2 = 4b^2 + 4a^2$ Now, we need to rearrange this equation to isolate $a^2$ and $b^2$ terms and find a direct relationship between them. Subtract $4b^2$ from both sides: $12b^2 + a^2 = 4a^2$ Subtract $a^2$ from both sides: $12b^2 = 3a^2$ Finally, divide both sides by 3 to simplify the relationship: $4b^2 = a^2$ This is a critical intermediate result! It tells us that $a^2$ is four times $b^2$. This relationship is all we need to determine the eccentricity.

5. Determine the Eccentricity ($e$)

From the relationship $a^2 = 4b^2$, we can clearly see that $a^2$ is greater than $b^2$. This directly implies that $a > b$. Since $a > b$, the major axis of the ellipse lies along the x-axis, and $a$ is the length of the semi-major axis. For this case ($a > b$), the correct formula for eccentricity is: $b^2 = a^2(1 - e^2)$ Now, we substitute the relationship $a^2 = 4b^2$ into this eccentricity formula. This allows us to eliminate $a^2$ and solve for $e^2$. $b^2 = (4b^2)(1 - e^2)$ Since $b^2$ cannot be zero (otherwise the points $(4,-1)$ and $(-2,2)$ wouldn't satisfy the ellipse equation, as it would imply division by zero or a degenerate ellipse), we can safely divide both sides of the equation by $b^2$: $1 = 4(1 - e^2)$ Divide both sides by 4: $\frac{1}{4} = 1 - e^2$ Rearrange the equation to solve for $e^2$: $e^2 = 1 - \frac{1}{4}$ $e^2 = \frac{4}{4} - \frac{1}{4}$ $e^2 = \frac{3}{4}$ Finally, take the square root of both sides to find $e$. Since eccentricity is always a positive value for an ellipse: $e = \sqrt{\frac{3}{4}}$ $e = \frac{\sqrt{3}}{\sqrt{4}}$ $e = \frac{\sqrt{3}}{2}$ This value of $e = \frac{\sqrt{3}}{2}$ (approximately 0.866) is indeed between 0 and 1, which is consistent for an ellipse.

Verification (Optional but Recommended for Understanding)

To ensure our calculations are correct, we can find the actual values of $a^2$ and $b^2$ and check if the given points lie on the resulting ellipse. Substitute $a^2 = 4b^2$ into Equation 1: $16b^2 + (4b^2) = (4b^2)b^2$ $20b^2 = 4b^4$ Since $b^2 \neq 0$, we can divide by $4b^2$: $5 = b^2$ Now, find $a^2$: $a^2 = 4b^2 = 4(5) = 20$. So, the specific equation of the ellipse is $\frac{x^2}{20} + \frac{y^2}{5} = 1$.

Let's check if the point $(4, -1)$ lies on it: $\frac{4^2}{20} + \frac{(-1)^2}{5} = \frac{16}{20} + \frac{1}{5} = \frac{4}{5} + \frac{1}{5} = \frac{5}{5} = 1$. (This is correct)

Let's check if the point $(-2, 2)$ lies on it: $\frac{(-2)^2}{20} + \frac{2^2}{5} = \frac{4}{20} + \frac{4}{5} = \frac{1}{5} + \frac{4}{5} = \frac{5}{5} = 1$. (This is also correct)

The successful verification confirms that our values for $a^2$ and $b^2$ are correct, and consequently, our eccentricity calculation is sound.

Tips for Success and Common Pitfalls

• Start with the Standard Equation: Always begin by writing the correct standard equation based on the given information about the ellipse's center and axis alignment. This sets the foundation for all subsequent steps.
• Systematic Substitution: Carefully substitute the coordinates of the given points into the ellipse equation. Double-check your arithmetic, especially with squares and negative signs.
• Algebraic Precision: Errors in algebraic manipulation (e.g., clearing denominators, combining terms, solving equations) are common pitfalls. Work slowly and verify each step.
• Identify Major Axis Correctly: Before applying the eccentricity formula, explicitly determine whether $a > b$ or $b > a$ from the relationship you found (e.g., $a^2 = 4b^2 \implies a > b$). Using the wrong formula for eccentricity is a frequent mistake.
• Eccentricity Range Check: For an ellipse, $0 < e < 1$. If your calculated $e$ falls outside this range (e.g., $e=0$ for a circle, $e=1$ for a parabola, or $e>1$ for a hyperbola), it's a strong indication of a calculation error.

Summary and Key Takeaway

To determine the eccentricity of an ellipse given its center, axis alignment, and points it passes through, the general procedure is as follows:

1. Write down the standard equation of the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
2. Substitute the coordinates of each given point into this equation to form a system of two algebraic equations involving $a^2$ and $b^2$.
3. Solve this system of equations