Given an ellipse $ D: \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1\ (a > 0,\ b > 0)$ which contains a circle $ C: x^2 \plus{} y^2 \equal{} 1.$ Draw a tangent line from the point $ P(0,\ b)$ on $ D$, let the extented line meets $ D$ at $ Q$. Draw a tangent line from $ Q$, let the extended line meets $ D$ at $ R$, draw a tangent line from $ R$, let the extended line meets $ D$ at $ P$. Express $ a$ as the function of $ b$.

Please translate the context into plain English.

Thanks in advance

kunny
1. **Identify the given equations and conditions:**
   - The ellipse \( D \) is given by the equation:
     \[
     \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
     \]
   - The circle \( C \) is given by the equation:
     \[
     x^2 + y^2 = 1
     \]
   - The point \( P \) on the ellipse is \( (0, b) \).

2. **Determine the tangent line at \( P \):**
   - The tangent line to the ellipse at \( P(0, b) \) can be found using the derivative of the ellipse equation. However, since \( P \) is on the y-axis, the tangent line is horizontal:
     \[
     y = b
     \]

3. **Find the coordinates of \( Q \):**
   - Let \( Q = (x_0, y_0) \) be the point where the tangent line from \( P \) intersects the ellipse again. Since \( P \) is on the y-axis and the tangent line at \( P \) is horizontal, \( Q \) must be symmetric about the y-axis. Thus, \( Q \) has coordinates \( (x_0, -b) \).

4. **Determine the tangent line at \( Q \):**
   - The tangent line at \( Q \) can be found using the point-slope form of the tangent to the ellipse:
     \[
     \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1
     \]
   - Substituting \( Q = (x_0, -b) \):
     \[
     \frac{x_0 x}{a^2} - \frac{b y}{b^2} = 1 \implies \frac{x_0 x}{a^2} - \frac{y}{b} = 1
     \]

5. **Find the coordinates of \( R \):**
   - Let \( R = (-x_0, y_0) \) be the point where the tangent line from \( Q \) intersects the ellipse again. By symmetry, \( R \) has coordinates \( (-x_0, y_0) \).

6. **Determine the tangent line at \( R \):**
   - The tangent line at \( R \) can be found similarly:
     \[
     -\frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1
     \]
   - Substituting \( R = (-x_0, y_0) \):
     \[
     -\frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1
     \]

7. **Find the relationship between \( a \) and \( b \):**
   - Since \( Q \) lies on the ellipse \( D \):
     \[
     \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1
     \]
   - Given \( y_0 = -1 \) (from the symmetry and the fact that \( QR \) is tangent to \( C \)):
     \[
     \frac{x_0^2}{a^2} + \frac{1}{b^2} = 1
     \]
   - The distance from the origin to the line \( PR \) is given by:
     \[
     \frac{x_0 b}{\sqrt{x_0^2 + (1 + b)^2}} = 1
     \]
   - Solving for \( x_0 \):
     \[
     x_0^2 = a^2 \left(1 - \frac{1}{b^2}\right)
     \]
   - Substituting \( x_0 \) back into the distance equation:
     \[
     \frac{a \sqrt{1 - \frac{1}{b^2}} b}{\sqrt{a^2 \left(1 - \frac{1}{b^2}\right) + (1 + b)^2}} = 1
     \]
   - Simplifying, we get:
     \[
     a = \frac{b(1 + b)}{b^2 - 1}
     \]

The final answer is \( \boxed{ a = \frac{b(1 + b)}{b^2 - 1} } \).