If the ellipse x²/a² + y²/b² = 1 meets the line x/7 + y/(2√6) = 1 on the x-axis and the line x/7 - y

Q: Wait, why did we assume the ellipse passes through these points? The question says it 'meets' the lines there.

Great question! In coordinate geometry, 'meeting' a line at a specific point means that point is a common solution to both equations. Since the point lies on the line, it must also satisfy the ellipse's equation. We are essentially using the lines to 'pinpoint' specific locations on the ellipse.

Q: Does the negative sign in the $y$-coordinate $(0, -2\sqrt{6})$ affect the value of $b^2$?

Not at all! When we substitute the point into the ellipse equation, we square the $y$-coordinate. Since $(-2\sqrt{6})^2 = 24$, the negative sign vanishes. The ellipse is symmetric, so it doesn't matter if the intercept is positive or negative; the distance from the center remains the same.

Q: How do I know if the ellipse is horizontal or vertical?

Look at your values for $a^2$ and $b^2$. If $a^2 > b^2$, the major axis is along the $x$-axis (horizontal). If $b^2 > a^2$, it's vertical. Here, $49 > 24$, so it's horizontal, and our formula $e = \sqrt{1 - \frac{b^2}{a^2}}$ is perfectly correct.

Question

If the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the line $\frac{x}{7} + \frac{y}{2\sqrt{6}} = 1$ on the x-axis and the line $\frac{x}{7} - \frac{y}{2\sqrt{6}} = 1$ on the y-axis, then the eccentricity of the ellipse is

Options:

(A) $\frac{5}{7}$

(B) $\frac{2\sqrt{6}}{7}$

(C) $\frac{3}{7}$

(D) $\frac{2\sqrt{5}}{7}$

Sigma Percentile · Accepted Answer

### The Geometry of Intersections

Imagine you are standing on the Cartesian plane, looking at an ellipse centered at the origin. It is a graceful, closed curve, defined by the equation:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

This ellipse is interacting with two specific lines given in the intercept form: $\frac{x}{7} + \frac{y}{2\sqrt{6}} = 1$ and $\frac{x}{7} - \frac{y}{2\sqrt{6}} = 1$. Our mission is to uncover the eccentricity of this ellipse, a measure of its 'flatness.'

### Decoding the Footprints

Let us look at the first line: $\frac{x}{7} + \frac{y}{2\sqrt{6}} = 1$. The problem states this line meets the ellipse on the $x$-axis, where the $y$-coordinate is zero.

Setting $y = 0$ in the line's equation, the $y$-term vanishes, leaving us with $\frac{x}{7} = 1$, which simplifies to $x = 7$. Thus, the ellipse passes through the point $(7, 0)$.

Now, consider the second line: $\frac{x}{7} - \frac{y}{2\sqrt{6}} = 1$. This line meets the ellipse on the $y$-axis, where the $x$-coordinate is zero.

Setting $x = 0$ in the line's equation, the $x$-term vanishes, leaving us with $-\frac{y}{2\sqrt{6}} = 1$. Solving for $y$, we get $y = -2\sqrt{6}$. So, the ellipse also passes through the point $(0, -2\sqrt{6})$.

### Determining the Parameters

We have two points on our ellipse: $(7, 0)$ and $(0, -2\sqrt{6})$. Let us substitute these into the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

For the point $(7, 0)$:

$$\frac{7^2}{a^2} + \frac{0^2}{b^2} = 1 \Rightarrow \frac{49}{a^2} = 1 \Rightarrow a^2 = 49$$

For the point $(0, -2\sqrt{6})$:

$$\frac{0^2}{a^2} + \frac{(-2\sqrt{6})^2}{b^2} = 1 \Rightarrow \frac{24}{b^2} = 1 \Rightarrow b^2 = 24$$

We have successfully extracted the parameters $a^2 = 49$ and $b^2 = 24$.

### The Final Calculation: Eccentricity

Since $a^2 > b^2$, we know our ellipse is stretched along the $x$-axis. The eccentricity $e$ is defined by the relationship $e = \sqrt{1 - \frac{b^2}{a^2}}$.

Substituting our values, we get:

$$e = \sqrt{1 - \frac{24}{49}}$$

Taking the common denominator, we find:

$$e = \sqrt{\frac{49 - 24}{49}} = \sqrt{\frac{25}{49}}$$

Taking the square root of the numerator and the denominator separately, we arrive at the final result:

**$$e = \frac{5}{7}$$**

The eccentricity is **$\frac{5}{7}$**. This problem is a beautiful reminder that even complex-looking geometry problems are just puzzles waiting to be solved by identifying the right points and applying fundamental definitions.

Animated Solution for Mathematics - Conic Sections: If the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ meets the line $\frac{x}{7} + \frac{y}{2 6} = 1$ on the x-axis and the line $\frac{x}{7} - \frac{y}{2 6} = 1$ on the y-axis, then the eccentricity of the ellipse is

Select Answer:

Visualized Solution

The Ellipse and the Lines

$x$ -axis Intercept of $L_{1}$

Finding $a^{2}$

$y$ -axis Intercept of $L_{2}$

Finding $b^{2}$

The Eccentricity Formula

Substituting $a^{2}$ and $b^{2}$

Calculating Eccentricity

Final Answer

The Sigma Insight: Foci, Directrices, and Eccentricity

The Geometry of Intersections

Decoding the Footprints

Determining the Parameters

The Final Calculation

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