The equation of the ellipse whose foci are `(pm5,0)` and one of its directrix is 5x= 36 is

Options

A. `(x^2)/(36)+(y^2)/(11)=1`
B. `(x^2)/(6)+(y^2)/(sqrt(11))=1`
C. `(x^2)/(6)+(y^2)/(11)=1`
D. None of these

Answer

To find the equation of the ellipse whose foci are at (±5, 0) and one of its directrices is given by the equation 5x = 36, we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The foci of the ellipse are located at (±c, 0), where c = 5. Therefore, we have:
- c = 5

### Step 2: Find the equation of the directrix
The directrix is given by the equation 5x = 36. We can rewrite this as:
- x = 36/5

### Step 3: Relate the parameters of the ellipse
For an ellipse, the relationship between the semi-major axis (a), semi-minor axis (b), and the distance to the foci (c) is given by:
- c² = a² - b²

Also, the distance from the center to the directrix is given by:
- Distance to directrix = a/e, where e is the eccentricity.

### Step 4: Find the eccentricity (e)
The distance from the center to the directrix (x = 36/5) is equal to:
- 36/5

Since the distance to the directrix is also given by a/e, we can set up the equation:
- a/e = 36/5

### Step 5: Express e in terms of a
From the above equation, we can express e as:
- e = 5a/36

### Step 6: Use the relationship between e, a, and c
The eccentricity (e) is also defined as:
- e = c/a

Substituting c = 5, we have:
- e = 5/a

### Step 7: Set the two expressions for e equal
Now we can set the two expressions for e equal:
- 5/a = 5a/36

### Step 8: Solve for a
Cross-multiplying gives us:
- 5 * 36 = 5a²
- 36 = a²
- a = 6

### Step 9: Find b using c² = a² - b²
Now that we have a, we can find b:
- c² = a² - b²
- 5² = 6² - b²
- 25 = 36 - b²
- b² = 36 - 25
- b² = 11

### Step 10: Write the equation of the ellipse
The standard form of the equation of the ellipse centered at the origin is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting a² and b²:
\[ \frac{x^2}{36} + \frac{y^2}{11} = 1 \]

### Final Result
The equation of the ellipse is:
\[ \frac{x^2}{36} + \frac{y^2}{11} = 1 \]

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