Suppose (h, k) lies on the circle `x^2 + y^2` =4, then (2h + 1, 3k + 2) lies on an ellipse with eccentricity e, then `e^2` equals

Answer

To solve the problem, we need to find the eccentricity \( e \) of the ellipse defined by the transformation of points from the circle \( x^2 + y^2 = 4 \).

### Step-by-Step Solution:

1. **Identify the Circle**: The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This means that the radius of the circle is \( r = 2 \) and it is centered at the origin \( (0, 0) \).

2. **Parameterize the Circle**: Any point \( (h, k) \) on the circle can be expressed in terms of a parameter \( \theta \):
\[ h = 2 \cos \theta, \quad k = 2 \sin \theta \]

3. **Transform the Point**: We need to find the new point \( (x', y') \) defined by:
\[ x' = 2h + 1 = 2(2 \cos \theta) + 1 = 4 \cos \theta + 1 \]
\[ y' = 3k + 2 = 3(2 \sin \theta) + 2 = 6 \sin \theta + 2 \]

4. **Express in Terms of \( x' \) and \( y' \)**: We can rewrite the expressions for \( x' \) and \( y' \):
\[ x' - 1 = 4 \cos \theta \quad \Rightarrow \quad \cos \theta = \frac{x' - 1}{4} \]
\[ y' - 2 = 6 \sin \theta \quad \Rightarrow \quad \sin \theta = \frac{y' - 2}{6} \]

5. **Use the Pythagorean Identity**: Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we substitute:
\[ \left(\frac{x' - 1}{4}\right)^2 + \left(\frac{y' - 2}{6}\right)^2 = 1 \]

6. **Simplify the Equation**: This leads to:
\[ \frac{(x' - 1)^2}{16} + \frac{(y' - 2)^2}{36} = 1 \]
This is the standard form of an ellipse.

7. **Identify \( a^2 \) and \( b^2 \)**: From the equation, we have:
\[ a^2 = 16 \quad \text{and} \quad b^2 = 36 \]

8. **Calculate the Eccentricity**: The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{16}{36}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

9. **Find \( e^2 \)**: Finally, we need \( e^2 \):
\[ e^2 = \left(\frac{\sqrt{5}}{3}\right)^2 = \frac{5}{9} \]

### Final Answer:
Thus, \( e^2 = \frac{5}{9} \).