The ellipse E1: x²/9 + y²/4 = 1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E2 is?

The ellipse E1 touches x-axis at (±3, 0) and y-axis at (0, ±2). Since the ellipse is inscribed in rectangle R whose sides are parallel to the coordinate axis, the vertices of rectangle are (±3, ±2).
Let the equation of ellipse E2 be x²/a² + y²/b² = 1.
The ellipse circumscribes the rectangle R, so the vertices of rectangle lie on ellipse E2. Therefore we get 9/a² + 4/b² = 1.
Given that ellipse E2 passes through (0, 4). So we get b = 4 and a² = 12.
We know that a² = b²(1 - e²)
⇒ e² = 1/4 ⇒ e = 1/2