Question

use implicit differentiation to find an equation of the tangent line to the cur
$x^{2}+6xy + 12y^{2}=28$, $(2,1)$ (ellipse)
$-\frac{1}{2}x + 2$

Explanation:

Step1: Differentiate both sides

Differentiate \(x^{2}+6xy + 12y^{2}=28\) with respect to \(x\).
Using the sum rule \((u + v+w)^\prime=u^\prime + v^\prime+w^\prime\), where \(u = x^{2}\), \(v=6xy\), \(w = 12y^{2}\).
For \(u=x^{2}\), \(u^\prime=2x\).
For \(v = 6xy\), use the product rule \((uv)^\prime=u^\prime v+uv^\prime\) (here \(u = 6x\), \(v = y\)), so \(v^\prime=6y+6x\frac{dy}{dx}\).
For \(w = 12y^{2}\), use the chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\), so \(w^\prime=24y\frac{dy}{dx}\).
The derivative of the right - hand side is \(0\).
So \(2x+6y + 6x\frac{dy}{dx}+24y\frac{dy}{dx}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Group the terms with \(\frac{dy}{dx}\):
\(6x\frac{dy}{dx}+24y\frac{dy}{dx}=-2x - 6y\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(6x + 24y)=-2x - 6y\).
Then \(\frac{dy}{dx}=\frac{-2x - 6y}{6x + 24y}=\frac{-x - 3y}{3x + 12y}\).

Step3: Evaluate \(\frac{dy}{dx}\) at the point \((2,1)\)

Substitute \(x = 2\) and \(y = 1\) into \(\frac{dy}{dx}\):
\(\frac{dy}{dx}\mid_{(x = 2,y = 1)}=\frac{-2-3\times1}{3\times2+12\times1}=\frac{-5}{18}\).

Step4: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=2\), \(y_{1}=1\), \(m =-\frac{5}{18}\).
\(y - 1=-\frac{5}{18}(x - 2)\).
Multiply through by \(18\): \(18y-18=-5x + 10\).
Rearrange to get \(5x+18y=28\).

Answer:

\(5x + 18y=28\)