If a ellipse has centre at (0,0), a focus at (-3,0) and the corresponding directrix is 3x + 25 = 0 , then it passes through the point :

Options

A. `(-5, -4)`
B. `(5/2, 4)`
C. `(-5, -4//sqrt(2))`
D. `(5//sqrt(2), 4//sqrt(2))`

Answer

To solve the problem step by step, we will derive the equation of the ellipse based on the given information and then check which point lies on the ellipse.

### Step 1: Identify the given information
- Center of the ellipse: \( (0, 0) \)
- Focus of the ellipse: \( (-3, 0) \)
- Directrix of the ellipse: \( 3x + 25 = 0 \) which simplifies to \( x = -\frac{25}{3} \)

### Step 2: Determine the values of \( a \) and \( e \)
For an ellipse centered at the origin with a horizontal major axis, the focus \( (ae, 0) \) corresponds to \( (-3, 0) \). Thus, we have:
\[ ae = 3 \quad \text{(1)} \]

The equation of the directrix is \( x = -\frac{a}{e} \). From the directrix equation:
\[ -\frac{a}{e} = -\frac{25}{3} \implies \frac{a}{e} = \frac{25}{3} \quad \text{(2)} \]

### Step 3: Solve equations (1) and (2)
From equation (1):
\[ e = \frac{3}{a} \]

Substituting \( e \) in equation (2):
\[ \frac{a}{\frac{3}{a}} = \frac{25}{3} \]
\[ \frac{a^2}{3} = \frac{25}{3} \implies a^2 = 25 \implies a = 5 \]

Now substituting \( a \) back into equation (1) to find \( e \):
\[ e = \frac{3}{5} \]

### Step 4: Find \( b^2 \)
Using the relationship \( e^2 = 1 - \frac{b^2}{a^2} \):
\[ \left(\frac{3}{5}\right)^2 = 1 - \frac{b^2}{25} \]
\[ \frac{9}{25} = 1 - \frac{b^2}{25} \]
\[ \frac{b^2}{25} = 1 - \frac{9}{25} = \frac{16}{25} \implies b^2 = 16 \]

### Step 5: 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 \( a^2 = 25 \) and \( b^2 = 16 \):
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]

### Step 6: Check which point lies on the ellipse
We need to check the given points to see which one satisfies the ellipse equation.

1. **Point (-5, -4)**:
\[ \frac{(-5)^2}{25} + \frac{(-4)^2}{16} = \frac{25}{25} + \frac{16}{16} = 1 + 1 = 2 \quad \text{(not on the ellipse)} \]

2. **Point \(\left(\frac{5}{2}, 4\right)\)**:
\[ \frac{\left(\frac{5}{2}\right)^2}{25} + \frac{4^2}{16} = \frac{\frac{25}{4}}{25} + \frac{16}{16} = \frac{1}{4} + 1 = \frac{5}{4} \quad \text{(not on the ellipse)} \]

3. **Point (-5, -4)**:
\[ \frac{(-5)^2}{25} + \frac{(-4)^2}{16} = 2 \quad \text{(not on the ellipse)} \]

4. **Point \(\left(\frac{5}{\sqrt{2}}, \frac{4}{\sqrt{2}}\right)\)**:
\[ \frac{\left(\frac{5}{\sqrt{2}}\right)^2}{25} + \frac{\left(\frac{4}{\sqrt{2}}\right)^2}{16} = \frac{\frac{25}{2}}{25} + \frac{\frac{16}{2}}{16} = \frac{1}{2} + \frac{1}{2} = 1 \quad \text{(on the ellipse)} \]

### Conclusion
The point that lies on the ellipse is \(\left(\frac{5}{\sqrt{2}}, \frac{4}{\sqrt{2}}\right)\).