The length of the minor axis (along the y-axis) of an ellipse in the standard form is `4/(sqrt(3))` If this ellipse touches the line, `x+6y=8,` then its eccentricity is :

Options

A. `1/2sqrt(5/3)`
B. `1/2sqrt(11/3)`
C. `sqrt(5/6)`
D. `1/3sqrt(11/3)`

Answer

To solve the problem, we need to find the eccentricity of the ellipse given that the length of the minor axis is \( \frac{4}{\sqrt{3}} \) and that the ellipse touches the line \( x + 6y = 8 \).

### Step 1: Identify the parameters of the ellipse
The standard form of an ellipse centered at the origin with the minor axis along the y-axis is given by:

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

where \( 2b \) is the length of the minor axis. Given that the length of the minor axis is \( \frac{4}{\sqrt{3}} \), we can find \( b \):

\[ 2b = \frac{4}{\sqrt{3}} \implies b = \frac{2}{\sqrt{3}} \]

### Step 2: Find the semi-major axis \( a \)
The relationship between the semi-major axis \( a \), semi-minor axis \( b \), and eccentricity \( e \) is given by:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

To find \( a \), we need to determine the distance from the center of the ellipse to the line \( x + 6y = 8 \).

### Step 3: Calculate the distance from the center to the line
The distance \( d \) from the center of the ellipse (which is at the origin (0,0)) to the line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For the line \( x + 6y - 8 = 0 \), we have \( A = 1, B = 6, C = -8 \). Thus, the distance \( d \) is:

\[ d = \frac{|1 \cdot 0 + 6 \cdot 0 - 8|}{\sqrt{1^2 + 6^2}} = \frac{8}{\sqrt{1 + 36}} = \frac{8}{\sqrt{37}} \]

### Step 4: Set the distance equal to the semi-major axis
Since the ellipse touches the line, the distance from the center to the line is equal to the semi-major axis \( a \):

\[ a = \frac{8}{\sqrt{37}} \]

### Step 5: Calculate \( b^2 \) and \( a^2 \)
Now we can calculate \( b^2 \) and \( a^2 \):

\[ b^2 = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]

\[ a^2 = \left(\frac{8}{\sqrt{37}}\right)^2 = \frac{64}{37} \]

### Step 6: Find the eccentricity \( e \)
Now we can substitute \( a^2 \) and \( b^2 \) into the eccentricity formula:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{\frac{4}{3}}{\frac{64}{37}}} \]

Calculating the fraction:

\[ \frac{b^2}{a^2} = \frac{4}{3} \cdot \frac{37}{64} = \frac{148}{192} = \frac{37}{48} \]

Thus,

\[ e = \sqrt{1 - \frac{37}{48}} = \sqrt{\frac{48 - 37}{48}} = \sqrt{\frac{11}{48}} = \frac{\sqrt{11}}{4\sqrt{3}} \]

### Final Answer
The eccentricity of the ellipse is:

\[ e = \frac{\sqrt{11}}{4\sqrt{3}} \]