#### 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 2(x2+y2)2=25(x2−y2) at the point (3,1) is y=−913x+4013.

#### Explanation

**Given:**

The curve is 2(x2+y2)2=25(x2−y2).

The point is (3,1).

**Derivative rule:** Chain rule

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydu⋅dudx.

**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,y)=(x1,y1).

**Calculation:**

Consider the given equation 2(x2+y2)2=25(x2−y2).

Differentiate the given equation implicitly with respect to *x*.

ddx(2(x2+y2)2)=ddx(25(x2−y2))2ddx(x2+y2)2=25ddx(x2−y2)

Let u=x2+y2 and k=x2−y2.

Then, 2ddx(u2)=25ddx(v).

Apply the chain rule and simplify the terms.

2[ddu(u2)dudx]=25[ddv(v)dvdx]2[2ududx]=25[1dvdx]4ududx=25dvdx

Substitute u=x2+y2 and k=x2−y2.

4(x2+y2)ddx(x2+y2)=25ddx(x2−y2)4(x2+y2)[ddx(x2)+ddx(y2)]=25[ddx(x2)−ddx(y2)]4(x2+y2)[2x+ddx(y2)]=25[2x−ddx(y2)]

Apply the chain rule and simplify the terms.

4(x2+y2)[2x+2ydydx]=25[2x−2ydydx]4(x2+y2)[x+ydydx]=25[x−ydydx]4x(x2+y2)+4y(x2+y2)dydx=25x−25ydydx

Combine the terms dydx to one side of the equation.

4y(x2+y2)dydx+25ydydx=25x−4x(x2+y2)(4y(x2+y2)+25y)dydx=25x−4x(x2+y2)dydx=25x−4x(x2+y2)4y(x2+y2)+25y

Therefore, the derivative of the equation is dydx=25x−4x(x2+y2)4y(x2+y2)+25y.

The slope of the tangent line at (3,1) is computed as follows,

dydx|(x,y)=(3,1)=25(3)−4(3)(32+12)4(1)(32+12)+25(1)=75−12(10)4(10)+25=−4565=−913

Thus, the slope of the tangent line at the point (3,1) is m=−913.

Substitute (3,1) for (x1,y1) and m=−913 in equation (1).

y−1=−913(x−3)y−1=−913x+2713y=−913x+2713+1y=−913x+4013

Therefore, the equation of the tangent line to the equation 2(x2+y2)2=25(x2−y2) at the point (3,1) is y=−913x+4013.

**Graph:**

The graph of the given curve and tangent line is shown below Figure 1.

From Figure 1, it is observed that the line y=−913x+4013 is tangent to the equation 2(x2+y2)2=25(x2−y2) at the point at (3,1).