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

To find the second point where the given curve has a horizontal tangent line, we start by implicitly differentiating the given equation \[ 25 x^{2} - 150 x + 9 y^{2} - 36 y = -36. \] Taking the derivative with respect to \( x \): \[ 50 x - 150 + 18 y \frac{dy}{dx} - 36 \frac{dy}{dx} = 0. \] This simplifies to: \[ \frac{dy}{dx} (18y - 36) = 150 - 50x. \] Now, for the tangent to be horizontal, \( \frac{dy}{dx} \) must equal zero. Thus, we set: \[ 18y - 36 = 0 \Rightarrow y = 2. \] To find the corresponding \( x \)-coordinates, substitute \( y = 2 \) back into the original equation: \[ 25 x^{2} - 150 x + 9(2^{2}) - 36(2) = -36, \] which simplifies to: \[ 25 x^{2} - 150 x + 36 - 72 = -36 \] or \[ 25 x^{2} - 150 x + 36 = -36. \] Rearranging gives: \[ 25 x^{2} - 150 x + 72 = 0. \] Dividing through by 25 leads to: \[ x^{2} - 6x + \frac{72}{25} = 0. \] Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \): \[ x = \frac{6 \pm \sqrt{36 - \frac{288}{25}}}{2}. \] Calculating the discriminant: \[ 36 - \frac{288}{25} = \frac{900 - 288}{25} = \frac{612}{25}. \] Now solving for \( x \): \[ x = \frac{6 \pm \frac{\sqrt{612}}{5}}{2} = \frac{30 \pm \sqrt{612}}{10}. \] Using the approximate value for \( \sqrt{612} \) (which is about 24.74), we get: \[ x \approx 5.47 \text{ and } 0.53. \] Since we are looking for the second point and already know one point is \( (3, 7) \), we can check an approximate value more accurately. So the result is: \( x \)-coordinate of second point \( = 0.53 \) \( y \)-coordinate of second point \( = 2 \)