Equation of ellipse having letus rectum 8 and eccentricity `(1)/(sqrt(2))` is

Options

A. `(x^(2))/(18) + (y^(2))/(32) = 1`
B. `(x^(2))/(8)+ (y^(2))/(9) = 1`
C. `(x^(2))/(64) + (y^(2))/(32) = 1`
D. `(x^(2))/(16) + (y^(2))/(24) = 1`

Answer

To find the equation of the ellipse with a given latus rectum and eccentricity, we can follow these steps:

### Step 1: Understand the properties of the ellipse
The standard form of the equation of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 2: Use the given latus rectum
The latus rectum \( L \) of an ellipse is given by the formula:
\[ L = \frac{2b^2}{a} \]
Given that the latus rectum is 8, we can set up the equation:
\[ \frac{2b^2}{a} = 8 \]
From this, we can simplify to find:
\[ b^2 = 4a \]

### Step 3: Use the given eccentricity
The eccentricity \( e \) of an ellipse is defined as:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Given that the eccentricity is \( \frac{1}{\sqrt{2}} \), we can square both sides:
\[ \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{b^2}{a^2} \]
This simplifies to:
\[ \frac{1}{2} = 1 - \frac{b^2}{a^2} \]
Rearranging gives:
\[ \frac{b^2}{a^2} = \frac{1}{2} \]

### Step 4: Substitute \( b^2 \) from Step 2 into Step 3
From Step 2, we have \( b^2 = 4a \). Substituting this into the equation from Step 3:
\[ \frac{4a}{a^2} = \frac{1}{2} \]
This simplifies to:
\[ \frac{4}{a} = \frac{1}{2} \]

### Step 5: Solve for \( a \)
Cross-multiplying gives:
\[ 4 \cdot 2 = a \implies a = 8 \]

### Step 6: Find \( b^2 \)
Using the value of \( a \) in the equation \( b^2 = 4a \):
\[ b^2 = 4 \cdot 8 = 32 \]

### Step 7: Write the equation of the ellipse
Now we can substitute \( a^2 \) and \( b^2 \) into the standard form of the ellipse:
\[ \frac{x^2}{8^2} + \frac{y^2}{32} = 1 \]
This simplifies to:
\[ \frac{x^2}{64} + \frac{y^2}{32} = 1 \]

### Final Answer
The equation of the ellipse is:
\[ \frac{x^2}{64} + \frac{y^2}{32} = 1 \]