The curve below has a horizontal tangent line at the point \( (3,7) \) and at one other point. Find the coordinates of the second point where the curve has a horizontal tangent line. \( 25 x^{2}-150 x+9 y^{2}-36 y=-36 \) \( x \)-coordinate of second point \( = \) \( y \)-coordinate of second point \( = \) Submit Question

**Step 1. Implicit Differentiation** The given curve is \[ 25x^2 - 150x + 9y^2 - 36y = -36. \] Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}(25x^2) - \frac{d}{dx}(150x) + \frac{d}{dx}(9y^2) - \frac{d}{dx}(36y) = \frac{d}{dx}(-36). \] This gives \[ 50x - 150 + 18y \frac{dy}{dx} - 36 \frac{dy}{dx} = 0. \] Factor \(\frac{dy}{dx}\) in the terms involving \( y \): \[ 50x - 150 + (18y - 36) \frac{dy}{dx} = 0. \] Solve for \(\frac{dy}{dx}\): \[ (18y - 36) \frac{dy}{dx} = 150 - 50x, \] so \[ \frac{dy}{dx} = \frac{150 - 50x}{18y - 36}. \] --- **Step 2. Horizontal Tangent Condition** A horizontal tangent occurs when \(\frac{dy}{dx} = 0\). The fraction is zero when the numerator is zero (and the denominator is not zero). Therefore, set \[ 150 - 50x = 0. \] Solve for \( x \): \[ 50x = 150 \quad \Longrightarrow \quad x = 3. \] Thus, any point on the curve that has a horizontal tangent must have \( x = 3 \). --- **Step 3. Find the Corresponding \( y \)-Coordinates** Substitute \( x = 3 \) into the original equation to find the corresponding \( y \)-values: \[ 25(3)^2 - 150(3) + 9y^2 - 36y = -36. \] Calculate \( 25(3)^2 \) and \( 150(3) \): \[ 25 \cdot 9 = 225, \quad 150 \cdot 3 = 450. \] Substitute these values: \[ 225 - 450 + 9y^2 - 36y = -36. \] Simplify the constant terms: \[ -225 + 9y^2 - 36y = -36. \] Bring \(-36\) to the left side: \[ 9y^2 - 36y - 225 + 36 = 0, \] which simplifies to \[ 9y^2 - 36y - 189 = 0. \] Divide the entire equation by 9: \[ y^2 - 4y - 21 = 0. \] Solve the quadratic equation by factoring or using the quadratic formula. The quadratic factors as \[ (y - 7)(y + 3) = 0. \] Thus, the solutions are \[ y = 7 \quad \text{or} \quad y = -3. \] --- **Step 4. Identify the Second Point** We are given that one of the points is \((3,7)\). Therefore, the second point where the curve has a horizontal tangent is \[ (3,-3). \] **Answer:** \( x \)-coordinate of second point \( = 3 \) \( y \)-coordinate of second point \( = -3 \)