#### To determine

**To find:** The equation of the tangent line to the given equation at the point.

#### Answer

The equation of the tangent line to the equation x2+2xy+4y2=12 at (2,1) is y=−x2+2.

#### Explanation

**Given:**

The curve is x2+2xy+4y2=12.

The point is (2,1).

**Derivative rules:**

(1) Chain rule: If y=f(u) and u=g(x) are both differentiable function, then

dydx=dydu⋅dudx.

(2) Product rule: If y=f(u) and u=g(x) are both differentiable function, then

ddx(f(x)+g(x))=ddx(f(x))+ddx(g(x)).

**Formula used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

where, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1,y=y1.

**Calculation:**

Consider the equation x2+2xy+4y2=12.

Differentiate the above equation implicitly with respect to *x*,

ddx(x2+2xy+4y2)=ddx(12)ddx(x2)+ddx(2xy)+ddx(4y2)=0ddx(x2)+2ddx(xy)+4ddx(y2)=0

Apply the product rule (2),

2x+2[xddx(y)+yddx(x)]+4ddx(y2)=02x+2[xdydx+y(1)]+4ddx(y2)=0

Apply the chain rule (1) and simplify the terms,

2x+2xdydx+2y+4[ddy(y2)dydx]=02x+2xdydx+2y+4[2ydydx]=02x+2xdydx+2y+8ydydx=0

Combine the terms dydx,

8ydydx+2xdydx=−2x−2ydydx(8y+2x)=−2x−2ydydx=−2x+2y8y+2xdydx=−x+y4y+x

Therefore, the derivative of the equation is dydx=−x+y4y+x.

The slope of the tangent line at (2,1) is computed as follows,

m=dydx|(x1,y1)=(2,1)=−2+14(1)+2=−36=−12

Thus, the slope of the tangent line at (2,1) is m=−12.

Substitute (2,1) for (x1,y1) and m=−12 in equation (1),

y−1=−12(x−2)y−1=−x2+1y=−x2+1+1y=−x2+2

Therefore, the equation of the tangent line to the equation x2+2xy+4y2=12 at (2,1) is y=−x2+2.