Question

Find those points on the curve \[ \frac{x^2}{4} + \frac{y^2}{25} = 1, \] on which the normal is parallel to the x-axis.

Hint:

For the normal to be parallel to the x-axis, the slope of the tangent must be 0. This occurs when \( x = 0 \) for this ellipse.

Answer 1

The equation of the ellipse is: \[ \frac{x^2}{4} + \frac{y^2}{25} = 1. \] To find the points where the normal is parallel to the x-axis, we first find the derivative of the equation to determine the slope of the tangent line.

Step 1: Implicit Differentiation. Differentiate both sides of the equation with respect to \( x \): \[ \frac{d}{dx} \left( \frac{x^2}{4} + \frac{y^2}{25} \right) = \frac{d}{dx}(1). \] This gives: \[ \frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0. \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{25x}{50y} = -\frac{x}{2y}. \]

Step 2: Condition for the normal being parallel to the x-axis. The normal is parallel to the x-axis if the slope of the normal is 0. The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, we need: \[ \frac{dy}{dx} = 0. \] This implies: \[ -\frac{x}{2y} = 0. \] Thus, \( x = 0 \).

Step 3: Find the corresponding value of \( y \). Substitute \( x = 0 \) into the original equation: \[ \frac{0^2}{4} + \frac{y^2}{25} = 1 $\Rightarrow$ \frac{y^2}{25} = 1 $\Rightarrow$ y^2 = 25 $\Rightarrow$ y = \pm 5. \] Conclusion: The points on the curve where the normal is parallel to the x-axis are \( (0, 5) \) and \( (0, -5) \).