The ends of 20 cm rope are at two points 16 cm apart. Find the eccentricity and latus rectum of ellipse formed by the variablepoint at which the rope is tighten.

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the Problem
We have a rope of length 20 cm with its ends at points A and B, which are 16 cm apart. The variable point P, where the rope is tightened, traces out an ellipse as it moves.

### Step 2: Identify the Foci and the Length of the Rope
Let the points A and B be the foci of the ellipse. The total length of the rope (PA + PB) is given as 20 cm. The distance between the foci (AB) is given as 16 cm.

### Step 3: Relate the Length of the Rope to the Ellipse
For an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant and equals 2a, where a is the semi-major axis. Thus:
\[ PA + PB = 20 \implies 2a = 20 \implies a = 10 \text{ cm} \]

### Step 4: Calculate the Distance Between the Foci
The distance between the foci (AB) is given as 16 cm. For an ellipse, the distance between the foci is given by:
\[ 2ae \quad \text{(where e is the eccentricity)} \]
Thus:
\[ 2ae = 16 \implies ae = 8 \]

### Step 5: Find the Eccentricity
Now we know \( a = 10 \) cm, we can find the eccentricity \( e \):
\[ e = \frac{8}{10} = \frac{4}{5} \]

### Step 6: Use the Eccentricity to Find b
We can use the relationship between a, b, and e in an ellipse:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the known values:
\[ \frac{4}{5} = \sqrt{1 - \frac{b^2}{10^2}} \implies \frac{16}{25} = 1 - \frac{b^2}{100} \]
Rearranging gives:
\[ \frac{b^2}{100} = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25} \]
Thus:
\[ b^2 = 100 \cdot \frac{9}{25} = 36 \implies b = 6 \text{ cm} \]

### Step 7: Find the Length of the Latus Rectum
The length of the latus rectum \( L \) of an ellipse is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values of \( b \) and \( a \):
\[ L = \frac{2 \cdot 36}{10} = \frac{72}{10} = \frac{36}{5} \text{ cm} \]

### Final Answers
- Eccentricity \( e = \frac{4}{5} \)
- Length of the latus rectum \( L = \frac{36}{5} \) cm