Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3, – 4), one focus at (4, – 4) and one vertex at (5, – 4). If mx – y = 4, `m gt 0` is a tangent to the ellipse E, then the value of `5 m^(2)` is equal to _____

Answer

To solve the problem, we need to find the value of \(5m^2\) given the conditions of the ellipse and the tangent line.

### Step-by-Step Solution:

1. **Identify the center, focus, and vertex of the ellipse**:
- The center of the ellipse \(E\) is at \((3, -4)\).
- One focus is at \((4, -4)\) and one vertex is at \((5, -4)\).

2. **Determine the values of \(a\) and \(c\)**:
- The distance from the center to the vertex (horizontal distance) is \(a\).
- Since the vertex is at \((5, -4)\) and the center is at \((3, -4)\), we have:
\[ a = 5 - 3 = 2 \]
- The distance from the center to the focus (horizontal distance) is \(c\).
- Since the focus is at \((4, -4)\) and the center is at \((3, -4)\), we have:
\[ c = 4 - 3 = 1 \]

3. **Calculate \(b\)**:
- We know the relationship between \(a\), \(b\), and \(c\) for an ellipse:
\[ c^2 = a^2 - b^2 \]
- Substituting the values of \(a\) and \(c\):
\[ 1^2 = 2^2 - b^2 \implies 1 = 4 - b^2 \implies b^2 = 3 \]

4. **Write the equation of the ellipse**:
- The standard form of the ellipse centered at \((h, k)\) is:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
- Substituting \(h = 3\), \(k = -4\), \(a^2 = 4\), and \(b^2 = 3\):
\[ \frac{(x - 3)^2}{4} + \frac{(y + 4)^2}{3} = 1 \]

5. **Find the equation of the tangent line**:
- The equation of the tangent line to the ellipse can be expressed as:
\[ y + 4 = m(x - 3) \pm \sqrt{a^2m^2 + b^2} \]
- Rearranging gives:
\[ y = mx - 3m - 4 \pm \sqrt{4m^2 + 3} \]

6. **Set the tangent line equal to the given line**:
- The given tangent line is \(mx - y = 4\) or \(y = mx - 4\).
- Equating the two expressions for \(y\):
\[ mx - 4 = mx - 3m - 4 \pm \sqrt{4m^2 + 3} \]
- This simplifies to:
\[ 0 = -3m \pm \sqrt{4m^2 + 3} \]

7. **Solve for \(m\)**:
- Setting the positive case:
\[ 3m = \sqrt{4m^2 + 3} \]
- Squaring both sides:
\[ 9m^2 = 4m^2 + 3 \implies 5m^2 = 3 \implies m^2 = \frac{3}{5} \]

8. **Calculate \(5m^2\)**:
- Finally, we find:
\[ 5m^2 = 3 \]

### Final Answer:
The value of \(5m^2\) is \(3\).