If radii of director circles of `x^2/a^2+y^2/b^2=1 and x^2/a^2-y^2/b^2=1` are `2r and r` respectively, let `e_E and e_H` are the eccentricities of ellipse and hyperbola respectively, then

Explanation

To solve the problem, we need to find the eccentricities of the ellipse and hyperbola given the radii of their director circles. Let's break it down step by step.

### Step 1: Identify the equations of the conics
The equations given are:
1. Ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
2. Hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)

### Step 2: Find the radius of the director circle of the ellipse
The radius of the director circle for the ellipse is given by:
\[ R_E = \sqrt{a^2 + b^2} \]
According to the problem, this radius is equal to \( 2r \):
\[ \sqrt{a^2 + b^2} = 2r \]
Squaring both sides gives:
\[ a^2 + b^2 = 4r^2 \quad \text{(1)} \]

### Step 3: Find the radius of the director circle of the hyperbola
The radius of the director circle for the hyperbola is given by:
\[ R_H = \sqrt{a^2 - b^2} \]
According to the problem, this radius is equal to \( r \):
\[ \sqrt{a^2 - b^2} = r \]
Squaring both sides gives:
\[ a^2 - b^2 = r^2 \quad \text{(2)} \]

### Step 4: Solve the equations (1) and (2)
Now we have two equations:
1. \( a^2 + b^2 = 4r^2 \) (from the ellipse)
2. \( a^2 - b^2 = r^2 \) (from the hyperbola)

We can solve these equations simultaneously.

Adding equations (1) and (2):
\[ (a^2 + b^2) + (a^2 - b^2) = 4r^2 + r^2 \]
This simplifies to:
\[ 2a^2 = 5r^2 \]
Thus,
\[ a^2 = \frac{5r^2}{2} \quad \text{(3)} \]

Now, substituting (3) into equation (1):
\[ \frac{5r^2}{2} + b^2 = 4r^2 \]
This gives:
\[ b^2 = 4r^2 - \frac{5r^2}{2} = \frac{8r^2 - 5r^2}{2} = \frac{3r^2}{2} \quad \text{(4)} \]

### Step 5: Find the eccentricities
The eccentricity \( e_E \) of the ellipse is given by:
\[ e_E = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting (3) and (4):
\[ e_E = \sqrt{1 - \frac{\frac{3r^2}{2}}{\frac{5r^2}{2}}} = \sqrt{1 - \frac{3}{5}} = \sqrt{\frac{2}{5}} = \frac{\sqrt{10}}{5} \]

The eccentricity \( e_H \) of the hyperbola is given by:
\[ e_H = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting (3) and (4):
\[ e_H = \sqrt{1 + \frac{\frac{3r^2}{2}}{\frac{5r^2}{2}}} = \sqrt{1 + \frac{3}{5}} = \sqrt{\frac{8}{5}} = \frac{2\sqrt{2}}{\sqrt{5}} \]

### Final Answer:
- Eccentricity of the ellipse \( e_E = \frac{\sqrt{10}}{5} \)
- Eccentricity of the hyperbola \( e_H = \frac{2\sqrt{2}}{\sqrt{5}} \)