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 of segment AB, where A and B are the points where a tangent to the ellipse \( 25x^2 + 9y^2 = 225 \) intersects the coordinate axes, given that \( OA = OB \).

### Step 1: Rewrite the equation of the ellipse
The given equation of the ellipse is:
\[ 25x^2 + 9y^2 = 225 \]
We can rewrite this in standard form by dividing by 225:
\[ \frac{x^2}{9} + \frac{y^2}{25} = 1 \]
This shows that the semi-major axis \( a = 5 \) and the semi-minor axis \( b = 3 \).

### Step 2: Identify the coordinates of points A and B
Let the coordinates of point A (where the tangent meets the x-axis) be \( (a, 0) \) and the coordinates of point B (where the tangent meets the y-axis) be \( (0, a) \). Since \( OA = OB \), we have \( OA = OB = a \).

### Step 3: Calculate the length of AB
The length of segment AB can be calculated using the distance formula. Since A and B form a right triangle with the origin O, we can use the Pythagorean theorem:
\[ AB = \sqrt{(a - 0)^2 + (0 - a)^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

### Step 4: Find the equation of the tangent to the ellipse
To find the equation of the tangent line to the ellipse at a point \( (3 \cos \theta, 5 \sin \theta) \), we use the formula for the tangent to the ellipse in intercept form:
\[ \frac{x}{3} + \frac{y}{5} = 1 \]

### Step 5: Find the x-intercept (A) and y-intercept (B)
1. **Finding the x-intercept (A)**:
Set \( y = 0 \):
\[ \frac{x}{3} + 0 = 1 \implies x = 3 \]
Thus, the x-intercept A is \( (3, 0) \).

2. **Finding the y-intercept (B)**:
Set \( x = 0 \):
\[ 0 + \frac{y}{5} = 1 \implies y = 5 \]
Thus, the y-intercept B is \( (0, 5) \).

### Step 6: Calculate the lengths OA and OB
Since \( OA = OB = a \), we have:
\[ OA = 3 \quad \text{and} \quad OB = 5 \]

### Step 7: Use the relationship to find A
From the intercepts, we have:
\[ A \cos \theta = 3 \quad \text{and} \quad A \sin \theta = 5 \]
Squaring and adding these equations:
\[ A^2 \cos^2 \theta + A^2 \sin^2 \theta = 3^2 + 5^2 \]
\[ A^2 (\cos^2 \theta + \sin^2 \theta) = 9 + 25 \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ A^2 = 34 \implies A = \sqrt{34} \]

### Step 8: Substitute A back into the length of AB
Now substituting \( A \) back into the length of AB:
\[ AB = A\sqrt{2} = \sqrt{34} \sqrt{2} = \sqrt{68} = 2\sqrt{17} \]

### Final Answer
Thus, the length \( AB \) is:
\[ \boxed{2\sqrt{17}} \]