12. [-/1 Points] DETAILS MY NOTES SCALCET9 3.5.032. Use implicit differentiation to find an equation of the tangent line to the curve at the given point. \[ x^{2}+6 x y+12 y^{2}=28, \quad(2,1) \quad \text { (ellipse) } \] Sten

To find the equation of the tangent line using implicit differentiation for the curve \(x^{2} + 6xy + 12y^{2} = 28\), you start by differentiating both sides with respect to \(x\). This yields: \[2x + 6\left(x\frac{dy}{dx} + y\right) + 24y\frac{dy}{dx} = 0.\] Now, solve for \(\frac{dy}{dx}\) at the point \((2, 1)\) by substituting \(x = 2\) and \(y = 1\). After some algebra, you'll get \(\frac{dy}{dx} = -\frac{2}{15}\). The equation of the tangent line can then be written in point-slope form, which gives you \(y - 1 = -\frac{2}{15}(x - 2)\). Now, solve for \(y\) to put it in slope-intercept form, and you’ll have your tangent line! When it comes to this curve, remembering the implicit differentiation rules is crucial, especially when dealing with mixed \(x\) and \(y\) terms. A common mistake is forgetting to apply the chain rule properly when differentiating \(y\). Always keep track of that \(\frac{dy}{dx}\) — it’s key to finding the slope!