#### 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 y2(y2−4)=x2(x2−5) at the point (0,−2) is y=−2.

#### Explanation

**Given:**

The curve is y2(y2−4)=x2(x2−5).

The point is (0,−2).

**Derivative rules:**

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

(2) Product rule: ddx(f(x)g(x))=f(x)ddx(g(x))+g(x)ddx(f(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:**

Obtain the equation of the tangent line to the equation at the point.

Rewrite the given equation as follows,

y2(y2−4)=x2(x2−5)y4−4y2=x4−5x2

Differentiate the equation implicitly with respect to *x*,

ddx(y4−4y2)=ddx(x4−5x2)ddx(y4)−4ddx(y2)=ddx(x4)−5ddx(x2)ddx(y4)−4ddx(y2)=4x2−10x

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

[ddy(y4)dydx]−4[ddy(y2)dydx]=4x2−10x4y3dydx−4[2ydydx]=4x2−10x4y3dydx−8ydydx=4x2−10x

Substitute (0,−2) for (x,y),

4(−2)3dydx−8(−2)dydx=4(0)2−10(0)(4)8dydx+16dydx=0dydx=0

Thus, the slope of the tangent at (0,−2) is m=0.

Substitute (0,−2) for (x1,y1) and m=0 in equation (1),

y+2=0(x−0)y+2=0y=−2

Therefore, the equation of the tangent line to the equation y2(y2−4)=x2(x2−5) at (0,−2) is y=−2.

**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=−2 is tangent to the equation y2(y2−4)=x2(x2−5) at the point (0,−2).