Axes are coordinate axes, A and L are the ends of major axis and latusrectum. Area of 1 `triangle` OAL = 8 sq. units, e =`(1)/sqrt(2)` then equation of the ellipse is

Options

A. `(x^(2))/(16)+(y^(2))/(8)=1`
B. `(x^(2))/(32)+(y^(2))/(16)=1`
C. `(x^(2))/(64)+(y^(2))/(32)=1`
D. `(x^(2))/(8)+(y^(2))/(4)=1`

Answer

To find the equation of the ellipse given the area of triangle OAL and the eccentricity, we can follow these steps:

### Step 1: Understand the given information
We are given:
- Area of triangle OAL = 8 sq. units
- Eccentricity \( e = \frac{1}{\sqrt{2}} \)

### Step 2: Set up the coordinates
Let:
- O (Origin) = (0, 0)
- A (End of Major Axis) = (0, b)
- L (Point on Latus Rectum) = (ae, b²/a)

### Step 3: Calculate the area of triangle OAL
The area \( A \) of triangle OAL can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( b \) (the y-coordinate of A) and the height is \( ae \) (the x-coordinate of L).

Thus, we have:
\[ 8 = \frac{1}{2} \times b \times ae \]
This simplifies to:
\[ b \cdot ae = 16 \quad \text{(Equation 1)} \]

### Step 4: Use the eccentricity to find a relationship between a and b
The eccentricity \( e \) is defined as:
\[ e = \frac{\sqrt{a^2 - b^2}}{a} \]
Substituting \( e = \frac{1}{\sqrt{2}} \):
\[ \frac{\sqrt{a^2 - b^2}}{a} = \frac{1}{\sqrt{2}} \]
Squaring both sides gives:
\[ \frac{a^2 - b^2}{a^2} = \frac{1}{2} \]
This leads to:
\[ 2(a^2 - b^2) = a^2 \quad \Rightarrow \quad 2a^2 - 2b^2 = a^2 \quad \Rightarrow \quad a^2 = 2b^2 \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
From Equation 2, we have:
\[ b \cdot a \cdot e = 16 \]
Substituting \( e = \frac{1}{\sqrt{2}} \):
\[ b \cdot a \cdot \frac{1}{\sqrt{2}} = 16 \quad \Rightarrow \quad b \cdot a = 16\sqrt{2} \]
Now substituting \( a = \sqrt{2}b \) from Equation 2:
\[ b \cdot \sqrt{2}b = 16\sqrt{2} \]
This simplifies to:
\[ 2b^2 = 16\sqrt{2} \quad \Rightarrow \quad b^2 = 8\sqrt{2} \]

### Step 6: Find a^2
Using Equation 2 again:
\[ a^2 = 2b^2 = 2(8\sqrt{2}) = 16\sqrt{2} \]

### Step 7: Write the equation of the ellipse
The standard form of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 \) and \( b^2 \):
\[ \frac{x^2}{16\sqrt{2}} + \frac{y^2}{8\sqrt{2}} = 1 \]
To simplify, we can multiply through by \( 16\sqrt{2} \):
\[ x^2 + 2y^2 = 32 \]
Thus, the equation of the ellipse is:
\[ \frac{x^2}{32} + \frac{y^2}{16} = 1 \]

### Final Answer:
The equation of the ellipse is:
\[ \frac{x^2}{32} + \frac{y^2}{16} = 1 \]