In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is Options: (A) $\frac{3}{5}$ (B) $\frac{1}{2}$ (C) $\frac{4}{5}$ (D) $\frac{1}{\sqrt{5}}$

### Analyzing the Setup We consider a standard horizontal ellipse centered at the origin, defined by the equation: $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ In this configuration, $a$ represents the semi-major axis along the $x$-axis, and $b$ represents the semi-minor axis along the $y$-axis. The eccentricity $e$ measures the deviation of the ellipse from a perfect circle. ### Decoding the Geometry The problem provides two critical geometric constraints. First, the distance between the foci is $6$. Since the foci are located at $(\pm ae, 0)$, the distance between them is $2ae$. $$ 2ae = 6 \Rightarrow ae = 3 $$ Second, the length of the minor axis is given as $8$. The minor axis spans from $(0, -b)$ to $(0, b)$, meaning its total length is $2b$. $$ 2b = 8 \Rightarrow b = 4 $$ ### The Bridge To find the eccentricity, we utilize the fundamental identity of the ellipse, which relates the semi-axes and the eccentricity: $$ b^2 = a^2(1 - e^2) $$ Expanding this expression, we obtain $b^2 = a^2 - a^2e^2$, which can be rewritten as $b^2 = a^2 - (ae)^2$. Substituting our known values $b = 4$ and $ae = 3$: $$ 4^2 = a^2 - 3^2 $$ $$ 16 = a^2 - 9 $$ $$ a^2 = 25 \Rightarrow a = 5 $$ ### Final Calculation With the values $a = 5$ and $ae = 3$ determined, we calculate the eccentricity $e$ using the ratio of the focal distance to the semi-major axis: $$ e = \frac{ae}{a} = \frac{3}{5} $$ The eccentricity of the ellipse is **$0.6$** or **$\frac{3}{5}$**. Since this value is strictly between $0$ and $1$, it confirms the geometric validity of our result.

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Animated Solution for Mathematics - Conic Sections: In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is

Q: Why is the distance between foci $2ae$ and not just $ae$?

The foci of a standard horizontal ellipse are located at $S(ae, 0)$ and $S'(-ae, 0)$. The distance between these two points is the difference in their x-coordinates, which is $ae - (-ae) = 2ae$. Always remember to account for both sides of the center!

Q: Why is $a$ always greater than $b$ in an ellipse?

In a standard horizontal ellipse, $a$ is the semi-major axis and $b$ is the semi-minor axis. By definition, the major axis is the longer one, so $a > b$. If you ever find $b > a$, it just means your ellipse is vertical, and the equation would be $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

Select Answer:

Visualized Solution

Visualizing the Ellipse

Let the equation of the ellipse be $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Here, $a$ is the semi-major axis and $b$ is the semi-minor axis.

Locating the Foci

The foci of a horizontal ellipse are located at $S (a e, 0)$ and $S^{'} (- a e, 0)$
Here, $e$ represents the eccentricity of the ellipse.

Distance Between Foci ( $2 a e$ )

The distance between the foci $S$ and $S^{'}$ is $2 a e$
Given: Distance between foci $= 6$
Therefore, we can write: $2 a e = 6$

Solving for $a e$

Divide both sides by $2$ :
$a e = \frac{6}{2} = 3$

Minor Axis Length ( $2 b$ )

The length of the minor axis is $2 b$
Given: Length of minor axis $= 8$
Therefore, we can write: $2 b = 8$

Solving for $b$

Divide both sides by $2$ :
$b = \frac{8}{2} = 4$

The Fundamental Identity

The fundamental relation for eccentricity in an ellipse is:
$b^{2} = a^{2} (1 - e^{2})$

Algebraic Transformation

Expanding the equation:
$b^{2} = a^{2} - a^{2} e^{2}$
We can rewrite this as:
$b^{2} = a^{2} - (a e)^{2}$

Substituting Known Values

Substitute $b = 4$ and $a e = 3$ into the equation:
$4^{2} = a^{2} - 3^{2}$

Calculating $a^{2}$

Simplify the squares:
$16 = a^{2} - 9$
Rearranging to solve for $a^{2}$ :
$a^{2} = 16 + 9 = 25$

Finding the Semi-Major Axis ( $a$ )

Taking the square root on both sides:
$a = 25 = 5$

Final Calculation of Eccentricity ( $e$ )

We have $a e = 3$ and $a = 5$
Therefore:
$e = \frac{a e}{a} = \frac{3}{5}$

Conclusion & Key Takeaway

Final Answer: Eccentricity $e = \frac{3}{5}$
Key Takeaway: The relation $b^{2} = a^{2} (1 - e^{2})$ is the bridge between axes and eccentricity.

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The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

We consider a standard horizontal ellipse centered at the origin, defined by the equation:

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

In this configuration,

a

represents the semi-major axis along the

x

-axis, and

b

represents the semi-minor axis along the

y

-axis. The eccentricity

e

measures the deviation of the ellipse from a perfect circle.

Decoding the Geometry

The problem provides two critical geometric constraints. First, the distance between the foci is

6

. Since the foci are located at

(\pm a e, 0)

, the distance between them is

2 a e

2 a e = 6 \Rightarrow a e = 3

Second, the length of the minor axis is given as

8

. The minor axis spans from

(0, - b)

(0, b)

, meaning its total length is

2 b

2 b = 8 \Rightarrow b = 4

The Bridge

To find the eccentricity, we utilize the fundamental identity of the ellipse, which relates the semi-axes and the eccentricity:

b^{2} = a^{2} (1 - e^{2})

Expanding this expression, we obtain

b^{2} = a^{2} - a^{2} e^{2}

, which can be rewritten as

b^{2} = a^{2} - (a e)^{2}

. Substituting our known values

b = 4

and

a e = 3

4^{2} = a^{2} - 3^{2}

16 = a^{2} - 9

a^{2} = 25 \Rightarrow a = 5

Final Calculation

With the values

a = 5

and

a e = 3

determined, we calculate the eccentricity

e

using the ratio of the focal distance to the semi-major axis:

e = \frac{a e}{a} = \frac{3}{5}

The eccentricity of the ellipse is $0.6$ or $\frac{3}{5}$ . Since this value is strictly between

0

and

1

, it confirms the geometric validity of our result.

Common Traps & FAQs

Why is the distance between foci

2 a e

and not just

a e

The foci of a standard horizontal ellipse are located at

S (a e, 0)

and

S^{'} (- a e, 0)

. The distance between these two points is the difference in their x-coordinates, which is

a e - (- a e) = 2 a e

. Always remember to account for both sides of the center!

Can I use

b^{2} = a^{2} (e^{2} - 1)

No, that formula is specifically for a hyperbola. For an ellipse, the eccentricity

e

is always between

0

and

1

, so

1 - e^{2}

must be positive. Always remember: Ellipse is

b^{2} = a^{2} (1 - e^{2})

, and Hyperbola is

b^{2} = a^{2} (e^{2} - 1)

Why is

a

always greater than

b

in an ellipse?

In a standard horizontal ellipse,

a

is the semi-major axis and

b

is the semi-minor axis. By definition, the major axis is the longer one, so

a > b

. If you ever find

b > a

, it just means your ellipse is vertical, and the equation would be

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

Animated Solution for Mathematics - Conic Sections: In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is

Select Answer:

Visualized Solution

Visualizing the Ellipse

Locating the Foci

Distance Between Foci ( $2 a e$ )

Solving for $a e$

Minor Axis Length ( $2 b$ )

Solving for $b$

The Fundamental Identity

Algebraic Transformation

Substituting Known Values

Calculating $a^{2}$

Finding the Semi-Major Axis ( $a$ )

Final Calculation of Eccentricity ( $e$ )

Conclusion & Key Takeaway

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Decoding the Geometry

The Bridge

Final Calculation

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Animated Solution for Mathematics - Conic Sections: In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is

Select Answer:

Visualized Solution

Visualizing the Ellipse

Locating the Foci

Distance Between Foci ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2ae)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ae

Minor Axis Length ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2b)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Fundamental Identity

Algebraic Transformation

Substituting Known Values

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Finding the Semi-Major Axis ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a)

Final Calculation of Eccentricity ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e)

Conclusion & Key Takeaway

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

Decoding the Geometry

The Bridge

Final Calculation

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