Key Concepts and Formulas

To master this problem, a strong understanding of the fundamental properties of an ellipse and linear equations is essential. We'll specifically focus on:

1. Standard Equation of an Ellipse: The most common form of an ellipse centered at the origin 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. Specifically, $a$ is the length of the semi-axis along the x-axis, and $b$ is the length of the semi-axis along the y-axis.
   • It's crucial to remember that $a$ and $b$ are always positive values, as they represent lengths.
   • This ellipse intersects the x-axis at the points $(\pm a, 0)$ and the y-axis at the points $(0, \pm b)$. These are called the x-intercepts and y-intercepts, respectively.

2. Intercept Form of a Line: A linear equation expressed as $\frac{x}{p}+\frac{y}{q}=1$ is known as the intercept form. This form directly gives us the points where the line crosses the coordinate axes:

   • The line intersects the x-axis at the point $(p, 0)$. Thus, $p$ is the x-intercept.
   • The line intersects the y-axis at the point $(0, q)$. Thus, $q$ is the y-intercept.

3. Eccentricity of an Ellipse ($e$): Eccentricity is a fundamental characteristic of an ellipse that quantifies its "ovalness" or "flatness." Its value $e$ always lies between $0$ and $1$ (i.e., $0 \le e < 1$). The formula for eccentricity depends on which semi-axis is the major axis (the longer one) and which is the minor axis (the shorter one).

   • If $a > b$ (meaning the major axis is along the x-axis), the eccentricity is given by: $e = \sqrt{1 - \frac{b^2}{a^2}}$
   • If $b > a$ (meaning the major axis is along the y-axis), the eccentricity is given by: $e = \sqrt{1 - \frac{a^2}{b^2}}$ A general way to remember this is $e = \sqrt{1 - \frac{(\text{semi-minor axis})^2}{(\text{semi-major axis})^2}}$. The term $\frac{(\text{semi-minor axis})^2}{(\text{semi-major axis})^2}$ is always less than 1, ensuring $e$ is real and between 0 and 1.

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Step-by-Step Solution

Our objective is to find the eccentricity of the given ellipse. To do this, we first need to determine the values of $a$ and $b$ from the provided conditions.

1. Understanding the Ellipse and its Intercepts

The given ellipse has the standard equation: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ From this form, we know that the ellipse crosses the x-axis at $(\pm a, 0)$ and the y-axis at $(0, \pm b)$. These points are crucial for connecting the ellipse to the given lines.

2. Using the First Condition to Determine the Value of 'a'

The problem states: "the ellipse meets the line $\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$ on the x-axis."

• Explanation of the condition: When the ellipse "meets the line on the x-axis," it means that their point of intersection lies on the x-axis. Any point on the x-axis has its y-coordinate equal to $0$.
• Step 2.1: Find the x-intercept of the given line. To find where the line $\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$ intersects the x-axis, we set $y=0$ in its equation. This is because all points on the x-axis have a y-coordinate of zero. $\frac{x}{7}+\frac{0}{2 \sqrt{6}}=1$ $\frac{x}{7}=1$ $x=7$ So, the line intersects the x-axis at the point $(7,0)$. This is the specific point where the ellipse and the line meet.
• Step 2.2: Relate this intercept to the ellipse's semi-axis 'a'. Since the ellipse meets this line at $(7,0)$ on the x-axis, this point must be one of the x-intercepts of the ellipse. We know the x-intercepts of the ellipse are $(\pm a, 0)$. By comparing $(7,0)$ with $(\pm a, 0)$, we can deduce: $a = 7$ We take the positive value because $a$ represents a length (the semi-axis length along the x-axis) and lengths are always positive.

3. Using the Second Condition to Determine the Value of 'b'

The problem states: "the ellipse meets the line $\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$ on the y-axis."

• Explanation of the condition: Similarly, "meets on the y-axis" means that their point of intersection lies on the y-axis. Any point on the y-axis has its x-coordinate equal to $0$.
• Step 3.1: Find the y-intercept of the given line. To find where the line $\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$ intersects the y-axis, we set $x=0$ in its equation. This is because all points on the y-axis have an x-coordinate of zero. $\frac{0}{7}-\frac{y}{2 \sqrt{6}}=1$ $-\frac{y}{2 \sqrt{6}}=1$ $y = -2\sqrt{6}$ So, the line intersects the y-axis at the point $(0, -2\sqrt{6})$. This is the specific point where the ellipse and the line meet.
• Step 3.2: Relate this intercept to the ellipse's semi-axis 'b'. Since the ellipse meets this line at $(0, -2\sqrt{6})$ on the y-axis, this point must be one of the y-intercepts of the ellipse. We know the y-intercepts of the ellipse are $(0, \pm b)$. By comparing $(0, -2\sqrt{6})$ with $(0, \pm b)$, we can deduce: $b = |-2\sqrt{6}| = 2\sqrt{6}$ Again, we take the absolute value because $b$ represents a length (the semi-axis length along the y-axis) and must be positive.

4. Calculate the Eccentricity of the Ellipse

Now we have the values for the semi-axes: $a=7$ and $b=2\sqrt{6}$. We can proceed to calculate the eccentricity $e$.

• Step 4.1: Identify the semi-major and semi-minor axes. Before applying the eccentricity formula, we need to compare $a$ and $b$ to determine which is the semi-major axis (the longer one). We have $a = 7$. For $b$, let's simplify and compare its squared value: $b = 2\sqrt{6} = \sqrt{4 \times 6} = \sqrt{24}$. Comparing $a$ and $b$: $a = 7 = \sqrt{49}$ $b = \sqrt{24}$ Since $\sqrt{49} > \sqrt{24}$, we have $a > b$. This means the semi-major axis is along the x-axis, and the semi-minor axis is along the y-axis.
• Step 4.2: Apply the correct eccentricity formula. Since $a > b$, we use the formula: $e = \sqrt{1 - \frac{b^2}{a^2}}$ Let's substitute the values $a=7$ and $b=2\sqrt{6}$: $e^2 = 1 - \frac{(2\sqrt{6})^2}{7^2}$ $e^2 = 1 - \frac{4 \times 6}{49}$ $e^2 = 1 - \frac{24}{49}$ To simplify, we find a common denominator: $e^2 = \frac{49}{49} - \frac{24}{49}$ $e^2 = \frac{49 - 24}{49}$ $e^2 = \frac{25}{49}$ Finally, take the square root to find $e$: $e = \sqrt{\frac{25}{49}}$ $e = \frac{\sqrt{25}}{\sqrt{49}}$ $e = \frac{5}{7}$ This value is between 0 and 1, which is consistent with the definition of eccentricity for an ellipse.

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Tips and Common Mistakes:

• Interpreting "Meets on the x-axis/y-axis": This phrase is key. It directly implies that the intersection point is an intercept of the line and an intercept of the ellipse. Remember that on the x-axis, $y=0$, and on the y-axis, $x=0$.
• Absolute Values for Axis Lengths: Always remember that $a$ and $b$ in the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ represent positive lengths of the semi-axes. If you calculate an intercept as negative (like $y = -2\sqrt{6}$), the corresponding axis length ($b$) is its absolute value ($2\sqrt{6}$).
• Choosing the Correct Eccentricity Formula: It is crucial to compare $a$ and $b$ first. The formula always involves $\frac{(\text{semi-minor axis})^2}{(\text{semi-major axis})^2}$. Swapping $a^2$ and $b^2$ in the denominator incorrectly will lead to an incorrect result (possibly $e^2 > 1$, which is impossible for an ellipse, or an imaginary value if you were to calculate $\sqrt{1 - \text{something greater than 1}}$).

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Summary and Key Takeaway

This problem is a straightforward application of the definitions of an ellipse's intercepts and eccentricity. By carefully interpreting the given conditions ("meets on the x-axis" and "meets on the y-axis"), we were able to directly determine the lengths of the semi-axes, $a=7$ and $b=2\sqrt{6}$. Once $a$ and $b$ were established, and we identified $a$ as the semi-major axis, we applied the standard eccentricity formula $e = \sqrt{1 - \frac{b^2}{a^2}}$. The final calculation yielded an eccentricity of $\frac{5}{7}$. The key takeaway is that geometric descriptions of a conic section often provide direct information about its defining parameters, which can then be used in formulas.

The correct option is (A).