If any tangent to the ellipse `25x^(2)+9y^2=225` meets the coordinate axes at A and B such that OA = OB then , the length AB is equal to (where , O is the origin)

Options

A. `sqrt17` units
B. `sqrt34` units
C. `2sqrt17` units
D. `2sqrt34` units

Answer

To solve the problem, we need to find the length \( AB \) where \( A \) and \( B \) are the points where the tangent to the ellipse \( 25x^2 + 9y^2 = 225 \) meets the coordinate axes, given that \( OA = OB \).

### Step-by-Step Solution:

1. **Rewrite the Equation of the Ellipse**:
The given ellipse equation is \( 25x^2 + 9y^2 = 225 \). We can rewrite it in standard form:
\[ \frac{x^2}{9} + \frac{y^2}{25} = 1 \]
Here, \( a^2 = 25 \) and \( b^2 = 9 \), so \( a = 5 \) and \( b = 3 \).

2. **General Equation of Tangent**:
The general equation of the tangent to the ellipse at point \( P(3 \cos \theta, 5 \sin \theta) \) is given by:
\[ \frac{x \cdot 3}{9} + \frac{y \cdot 5}{25} = 1 \]
Simplifying this, we get:
\[ \frac{x}{3} + \frac{y}{5} = 1 \]

3. **Finding Intercepts**:
To find the intercepts on the axes:
- For the x-intercept (where \( y = 0 \)):
\[ \frac{x}{3} = 1 \implies x = 3 \]
So, point \( A(3, 0) \).
- For the y-intercept (where \( x = 0 \)):
\[ \frac{y}{5} = 1 \implies y = 5 \]
So, point \( B(0, 5) \).

4. **Condition \( OA = OB \)**:
Given \( OA = OB \), we denote \( OA = OB = k \). We have:
\[ OA = 3 \quad \text{and} \quad OB = 5 \]
Since \( OA \) and \( OB \) must be equal, we set:
\[ k = OA = OB \]

5. **Finding Length \( AB \)**:
The length \( AB \) can be calculated using the distance formula:
\[ AB = \sqrt{(3 - 0)^2 + (0 - 5)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \]

6. **Final Length Calculation**:
Since \( OA = OB \), we can express the length \( AB \) in terms of \( k \):
\[ AB = OA + OB = k + k = 2k \]
Given \( OA = 3 \) and \( OB = 5 \), we find:
\[ k = \sqrt{34} \]
Thus, the length \( AB \) is:
\[ AB = 2 \sqrt{34} \]

### Conclusion:
The length \( AB \) is equal to \( 2\sqrt{17} \).