Chord of the ellipse `x^2/25+y^2/16=1` whose middle point is `(1/2, 2/5)` meets the minor axis at A and major axis at B, length of AB (in units) is :

Options

A. `sqrt41/5`
B. `(2sqrt41)/5`
C. `(3sqrt41)/5`
D. `(7sqrt41)/5`

Answer

To solve the problem, we need to find the length of the line segment AB, where A is the point where the chord meets the minor axis and B is the point where the chord meets the major axis of the ellipse defined by the equation \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \). The midpoint of the chord is given as \( \left(\frac{1}{2}, \frac{2}{5}\right) \).

### Step-by-Step Solution:

1. **Identify the parameters of the ellipse**:
The equation of the ellipse is given as:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]
Here, \( a^2 = 25 \) and \( b^2 = 16 \). Thus, \( a = 5 \) and \( b = 4 \). The major axis is along the x-axis and the minor axis is along the y-axis.

2. **Use the midpoint formula for the chord**:
The equation of a chord with midpoint \( (x_1, y_1) \) is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} \]
Substituting \( x_1 = \frac{1}{2} \) and \( y_1 = \frac{2}{5} \):
\[ \frac{xx_1}{25} + \frac{yy_1}{16} = \frac{\left(\frac{1}{2}\right)^2}{25} + \frac{\left(\frac{2}{5}\right)^2}{16} \]

3. **Calculate the right side of the equation**:
\[ \frac{\left(\frac{1}{2}\right)^2}{25} = \frac{\frac{1}{4}}{25} = \frac{1}{100} \]
\[ \frac{\left(\frac{2}{5}\right)^2}{16} = \frac{\frac{4}{25}}{16} = \frac{4}{400} = \frac{1}{100} \]
Thus, the right side becomes:
\[ \frac{1}{100} + \frac{1}{100} = \frac{2}{100} = \frac{1}{50} \]

4. **Set up the equation of the chord**:
Now, substituting back into the chord equation:
\[ \frac{xx_1}{25} + \frac{yy_1}{16} = \frac{1}{50} \]
This gives:
\[ \frac{x \cdot \frac{1}{2}}{25} + \frac{y \cdot \frac{2}{5}}{16} = \frac{1}{50} \]
Simplifying:
\[ \frac{x}{50} + \frac{2y}{80} = \frac{1}{50} \]
\[ \frac{x}{50} + \frac{y}{40} = \frac{1}{50} \]

5. **Find the points A and B**:
- **Point A (on the minor axis)**: Set \( x = 0 \):
\[ \frac{0}{50} + \frac{y}{40} = \frac{1}{50} \implies \frac{y}{40} = \frac{1}{50} \implies y = \frac{40}{50} = \frac{4}{5} \]
Thus, \( A = (0, \frac{4}{5}) \).

- **Point B (on the major axis)**: Set \( y = 0 \):
\[ \frac{x}{50} + \frac{0}{40} = \frac{1}{50} \implies \frac{x}{50} = \frac{1}{50} \implies x = 1 \]
Thus, \( B = (1, 0) \).

6. **Calculate the length of AB**:
Using the distance formula:
\[ AB = \sqrt{(1 - 0)^2 + \left(0 - \frac{4}{5}\right)^2} = \sqrt{1^2 + \left(-\frac{4}{5}\right)^2} = \sqrt{1 + \frac{16}{25}} = \sqrt{\frac{25}{25} + \frac{16}{25}} = \sqrt{\frac{41}{25}} = \frac{\sqrt{41}}{5} \]

### Final Answer:
The length of AB is \( \frac{\sqrt{41}}{5} \) units.