The point on the curve $\frac{x^2}{4} + \frac{y^2}{9} = 1$ at which the tangent to the curve is parallel to the x-axis is

The equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$ represents an ellipse.

This can be written as: $\frac{x^2}{2^2} + \frac{y^2}{3^2} = 1$

The ellipse is centered at the origin (0, 0), extends 2 units horizontally and 3 units vertically.

A tangent parallel to the x-axis is a horizontal line with slope = 0.

For an ellipse, horizontal tangents occur at the topmost and bottommost points.

These are the points where y reaches its maximum and minimum values.

From the equation $\frac{x^2}{2^2} + \frac{y^2}{3^2} = 1$:

The maximum value of y is 3 (top of ellipse)

The minimum value of y is -3 (bottom of ellipse)

At these extreme y-values, x = 0.

The points where the tangent is parallel to the x-axis are:

Top point: $(0, 3)$

Bottom point: $(0, -3)$

Therefore, the point is $(0, 3)$.