#### 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−xy−y2=1 at (2,1) is y=3x4−12.

#### Explanation

**Given:**

The equation is x2−xy−y2=1.

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−xy−y2=1.

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

ddx(x2−xy−y2)=ddx(1)ddx(x2)−ddx(xy)−ddx(y2)=0

Apply the product rule (2),

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

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

2x−[xdydx+y(1)]=ddy(y2)dydx2x−xdydx−y=2ydydxdydx=y−2x−x−2ydydx=2x−yx+2y

Therefore, the derivative of *y* is dydx=2x−yx+2y.

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

m=dydx|(x,y)=(2,1)=2(2)−12+2(1)=4−12+2=34

Thus, the slope of the tangent line at (2,1) is m=34.

Substitute (2,1) for (x1,y1) and m=34 in equation (1),

y−1=34(x−2)y−1=3x4−64y=3x4−32+1y=3x4−12

Therefore, the equation of the tangent line to the equation x2−xy−y2=1 at (2,1) is y=3x4−12.