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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

Ask by Salazar Valdez. in the United States
Mar 12,2025

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Tutor-Verified Answer

Answer

The second point where the curve has a horizontal tangent is at \( (3, -3) \).

Solution

**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 \( = -3 \)

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The Deep Dive

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 \( = 2 \)

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