Let's solve the problem step by step: #### Given Lemniscate Equation: \[ (x^2 + y^2)^2 - 4x^2 = 0 \] #### Step 1: Find \(\frac{dy}{dx}\) (Implicit Differentiation) Let \( F(x, y) = (x^2 + y^2)^2 - 4x^2 = 0 \). Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}\left((x^2 + y^2)^2 - 4x^2\right) = 0 \] Expand using the chain and product rule: \[ 2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx}) - 8x = 0 \] \[ 2(x^2 + y^2) [2x + 2y \frac{dy}{dx}] - 8x = 0 \] \[ 2(x^2 + y^2) \cdot 2x + 2(x^2 + y^2) \cdot 2y \frac{dy}{dx} - 8x = 0 \] \[ 4x(x^2 + y^2) + 4y(x^2 + y^2)\frac{dy}{dx} - 8x = 0 \] \[ 4y(x^2 + y^2)\frac{dy}{dx} = 8x - 4x(x^2 + y^2) \] \[ \frac{dy}{dx} = \frac{8x - 4x(x^2 + y^2)}{4y(x^2 + y^2)} \] \[ \frac{dy}{dx} = \frac{2x - x(x^2 + y^2)}{y(x^2 + y^2)} \] \[ \frac{dy}{dx} = \frac{x[2 - (x^2 + y^2)]}{y(x^2 + y^2)} \] --- #### Step 2: Set \(\frac{dy}{dx} = 0\) for Horizontal Tangent The tangent is horizontal when the numerator is 0: \[ x[2 - (x^2 + y^2)] = 0 \] So, \(x = 0\) **or** \(x^2 + y^2 = 2\). --- #### (A) **Case 1: \(x = 0\)** Plug into original equation: \[ (x^2 + y^2)^2 - 4x^2 = (0 + y^2)^2 - 0 = y^4 = 0 \implies y = 0 \] **So, one point: \((0, 0)\)** But let's double-check: Is this really a "point" on the curve? Substitute \(x=0, y=0\) into the original: \[ (0^2 + 0^2)^2 - 4 \cdot 0^2 = 0 - 0 = 0 \] So, yes, \((0,0)\) **is on the curve.** --- #### (B) **Case 2: \(x^2 + y^2 = 2\)** Plug into the original equation: \[ (x^2 + y^2)^2 - 4x^2 = 0 \] \[ (2)^2 - 4x^2 = 0 \] \[ 4 - 4x^2 = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1 \] For each \(x\), find corresponding \(y\): \[ x^2 + y^2 = 2 \] \[ 1 + y^2 = 2 \implies y^2 = 1 \implies y = 1 \text{ or } y = -1 \] #### So, the points are: 1. \(x = 1, y = 1\) 2. \(x = 1, y = -1\) 3. \(x = -1, y = 1\) 4. \(x = -1, y = -1\) --- ### **Final Answer:** **The coordinates of the four points at which the tangent line is horizontal are:** \[ \boxed{(1, 1),\ (1, -1),\ (-1, 1),\ (-1, -1)} \] And the origin \((0,0)\) is a degenerate case (a singular point), not a regular point with a tangent, so it is not included in the four regular points.

The tangent is horizontal where \(\frac{dy}{dx} = 0\), which occurs when the numerator \(x[2 - (x^2 + y^2)] = 0\). This gives points either with \(x=0\) or on the circle \(x^2 + y^2=2\). Substituting these into the curve's equation yields the points \((\pm1, \pm1)\), which are where the tangent line is horizontal.