The length of the semi-major axis of an ellipse is 8 and the eccentricity is `1/2`. If `Delta` denotes the area of the rectangle formed by joining the vertices of the latera recta of the ellipse then the value of `Delta` is equal to

Answer

To solve the problem, we need to find the area of the rectangle formed by joining the vertices of the latera recta of the ellipse. Let's go through the steps systematically.

### Step 1: Identify the parameters of the ellipse
Given:
- Length of the semi-major axis \( a = 8 \)
- Eccentricity \( e = \frac{1}{2} \)

### Step 2: Calculate the semi-minor axis \( b \)
The relationship between the semi-major axis, semi-minor axis, and eccentricity is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the known values:
\[ \frac{1}{2} = \sqrt{1 - \frac{b^2}{8^2}} = \sqrt{1 - \frac{b^2}{64}} \]
Squaring both sides:
\[ \left(\frac{1}{2}\right)^2 = 1 - \frac{b^2}{64} \]
\[ \frac{1}{4} = 1 - \frac{b^2}{64} \]
Rearranging gives:
\[ \frac{b^2}{64} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus,
\[ b^2 = 64 \cdot \frac{3}{4} = 48 \]

### Step 3: Find the coordinates of the vertices of the latera recta
The vertices of the latera recta of the ellipse are given by the points:
\[ \left( \pm ae, \frac{b^2}{a} \right) \]
Calculating \( ae \):
\[ ae = 8 \cdot \frac{1}{2} = 4 \]
Now, substituting \( b^2 \) and \( a \):
\[ \frac{b^2}{a} = \frac{48}{8} = 6 \]
So, the coordinates of the vertices of the latera recta are:
\[ (4, 6), (-4, 6), (4, -6), (-4, -6) \]

### Step 4: Calculate the area of the rectangle formed by these vertices
The length of the rectangle is the distance between the x-coordinates of the vertices:
\[ \text{Length} = 4 - (-4) = 8 \]
The width of the rectangle is the distance between the y-coordinates of the vertices:
\[ \text{Width} = 6 - (-6) = 12 \]
Thus, the area \( \Delta \) of the rectangle is given by:
\[ \Delta = \text{Length} \times \text{Width} = 8 \times 12 = 96 \]

### Final Answer
The value of \( \Delta \) is \( 96 \).

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