If a tangent to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` makes equal intercepts of length l on cordinates axes, then the values of l is

Options

A. `a^(2)+b^(2)`
B. `sqrt(a^(2)+b^(2))`
C. `(a^(2)+b^(2))^(2)`
D. None of these

Explanation

To solve the problem, we need to find the value of \( l \) such that a tangent to the ellipse

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

makes equal intercepts of length \( l \) on the coordinate axes.

### Step-by-step Solution:

1. **Equation of the Tangent to the Ellipse**:
The equation of the tangent to the ellipse at an angle \( \theta \) can be expressed as:
\[ \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \]

2. **Intercepts on the Axes**:
From the tangent equation, we can determine the intercepts:
- **X-intercept**: Set \( y = 0 \):
\[ \frac{x}{a} \cos \theta = 1 \implies x = \frac{a}{\cos \theta} \]
- **Y-intercept**: Set \( x = 0 \):
\[ \frac{y}{b} \sin \theta = 1 \implies y = \frac{b}{\sin \theta} \]

3. **Equal Intercepts Condition**:
According to the problem, the intercepts on both axes are equal and equal to \( l \):
\[ \frac{a}{\cos \theta} = l \quad \text{and} \quad \frac{b}{\sin \theta} = l \]

4. **Expressing Cosine and Sine**:
From the equal intercepts, we can express \( \cos \theta \) and \( \sin \theta \):
\[ \cos \theta = \frac{a}{l} \quad \text{and} \quad \sin \theta = \frac{b}{l} \]

5. **Using Pythagorean Identity**:
We know that:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Substituting the expressions for \( \cos \theta \) and \( \sin \theta \):
\[ \left(\frac{a}{l}\right)^2 + \left(\frac{b}{l}\right)^2 = 1 \]

6. **Simplifying the Equation**:
This simplifies to:
\[ \frac{a^2 + b^2}{l^2} = 1 \]

7. **Solving for \( l^2 \)**:
Rearranging gives:
\[ l^2 = a^2 + b^2 \]

8. **Finding \( l \)**:
Taking the square root of both sides, we find:
\[ l = \sqrt{a^2 + b^2} \]

### Final Answer:
Thus, the value of \( l \) is:
\[ l = \sqrt{a^2 + b^2} \]