Explanation: The equation of the given curve: $\frac{x^2}{9} + \frac{y^2}{25} = 1$On differentiating both sides w.r.t. $x$ , we get $\frac{2x}{9} + \frac{2y}{25} \frac{dy}{dx} = 1 \;\;\; \cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)$ $\frac{dy}{dx} = \frac{-25x}{9y}$Since, tangent is parallel to $X$ -axis, then the slope of the tangent is zero. $\therefore \;\; \frac{-25}{9} \frac{x}{y} = 0$ , which is possible if $x=0$Put $x=0$ in eq (i), we get $\frac{y^2}{25} = 1 \Rightarrow y^2 = 25 \Rightarrow y = \pm 5$Hence, required points are $(0, \pm 5)$ .