If radii of director circle of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` and hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` are in the ratio `1:3 and 4e_(1)^(2)-e_(2)^(2)=lambda`, where `e_1 and e_2` are the eccetricities of ellipse and hyperbola respectively, then the value of `lambda` is

Answer

To solve the problem, we need to find the value of \( \lambda \) given the equations of the ellipse and hyperbola, their director circles, and the relationship between their eccentricities. Let's break down the solution step by step.

### Step 1: Understand the equations of the ellipse and hyperbola
The equations given are:
- Ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
- Hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)

### Step 2: Find the radii of the director circles
The radius of the director circle for the ellipse is given by:
\[ R_1 = \sqrt{a^2 + b^2} \]
The radius of the director circle for the hyperbola is given by:
\[ R_2 = \sqrt{a^2 - b^2} \]

### Step 3: Set up the ratio of the radii
According to the problem, the ratio of the radii is \( 1:3 \):
\[ \frac{R_1}{R_2} = \frac{1}{3} \]
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 - b^2}} = \frac{1}{3} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ \frac{a^2 + b^2}{a^2 - b^2} = \frac{1}{9} \]

### Step 5: Cross-multiply to solve for \( a^2 \) and \( b^2 \)
Cross-multiplying results in:
\[ 9(a^2 + b^2) = a^2 - b^2 \]
Rearranging gives:
\[ 9a^2 + 9b^2 = a^2 - b^2 \]
\[ 8a^2 = -10b^2 \]
Thus, we can express \( a^2 \) in terms of \( b^2 \):
\[ a^2 = -\frac{5}{4}b^2 \]

### Step 6: Calculate the eccentricities
The eccentricity of the ellipse \( e_1 \) is given by:
\[ e_1 = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting \( a^2 = -\frac{5}{4}b^2 \):
\[ e_1^2 = 1 - \frac{b^2}{-\frac{5}{4}b^2} = 1 + \frac{4}{5} = \frac{9}{5} \]

The eccentricity of the hyperbola \( e_2 \) is given by:
\[ e_2 = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \( a^2 = -\frac{5}{4}b^2 \):
\[ e_2^2 = 1 + \frac{b^2}{-\frac{5}{4}b^2} = 1 - \frac{4}{5} = \frac{1}{5} \]

### Step 7: Use the relationship to find \( \lambda \)
We know:
\[ \lambda = 4e_1^2 - e_2^2 \]
Substituting the values of \( e_1^2 \) and \( e_2^2 \):
\[ \lambda = 4 \cdot \frac{9}{5} - \frac{1}{5} = \frac{36}{5} - \frac{1}{5} = \frac{35}{5} = 7 \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = 7 \]