Find all points on the curve \( 5 x y^{2}-25 x^{2} y=2 \) where it has a horizontal tangent line

To find the points on the curve \( 5xy^2 - 25x^2y = 2 \) where the tangent line is horizontal, we need to first calculate the derivative of the function implicitly. 1. Differentiate both sides of the equation with respect to \( x \): \[ \frac{d}{dx}(5xy^2) - \frac{d}{dx}(25x^2y) = \frac{d}{dx}(2) \] Using the product rule, we get: \[ 5\left(y^2 + 2xy\frac{dy}{dx}\right) - 25\left(2xy\frac{dy}{dx} + x^2\frac{dy}{dx}\right) = 0 \] Simplifying gives: \[ 5y^2 + 10xy\frac{dy}{dx} - 50xy\frac{dy}{dx} - 25x^2\frac{dy}{dx} = 0 \] Collecting like terms results in: \[ 5y^2 + \left(10xy - 50xy - 25x^2\right)\frac{dy}{dx} = 0 \] \[ 5y^2 - (40xy + 25x^2)\frac{dy}{dx} = 0 \] Thus, \[ \frac{dy}{dx} = \frac{5y^2}{40xy + 25x^2} \] 2. For the tangent line to be horizontal, \( \frac{dy}{dx} = 0 \). This occurs when the numerator is zero: \[ 5y^2 = 0 \implies y^2 = 0 \implies y = 0 \] 3. Substitute \( y = 0 \) back into the original curve equation to determine the values for \( x \): \[ 5x(0)^2 - 25x^2(0) = 2 \implies 0 = 2 \] This provides no valid values for \( x \). Therefore, the curve has no points where the tangent line is horizontal. However, double-checking the original equation shows there are no intersections where \( y = 0 \). So there are no points on the curve where the tangent line is horizontal—just a critical point to consider!