Chapter 3.5, Problem 28E

To determine

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

Answer to Problem 28E

The equation of the tangent line to the equation $y^2\left(y^2-4\right)=x^2\left(x^2-5\right)$ at the point $\left(0,-2\right)$ is $y=-2$.

Explanation of Solution

Given:

The curve is $y^2\left(y^2-4\right)=x^2\left(x^2-5\right)$.

The point is $\left(0,-2\right)$.

Derivative rules:

(1) Chain rule: If $y=f\left(u\right)$ and $u=g\left(x\right)$  are both differentiable function, then $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$.

(2) Product rule: $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=f\left(x\right)\frac{d}{dx}\left(g\left(x\right)\right)+g\left(x\right)\frac{d}{dx}\left(f\left(x\right)\right)$

Formula used:

The equation of the tangent line at $\left(x_1,y_1\right)$ is, $y-y_1=m\left(x-x_1\right)$ (1)

Where, m is the slope of the tangent line at $\left(x_1,y_1\right)$ and ${m=\frac{dy}{dx}|}_{x=x_1,y=y_1}$.

Calculation:

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

Rewrite the given equation as follows,

$\begin{array}{l}{y}^2\left(y^2-4\right)=x^2\left(x^2-5\right)\\ y^4-4y^2=x^4-5x^2\end{array}$

Differentiate the equation implicitly with respect to x,

$\begin{array}{l}\frac{d}{dx}\left(y^4-4y^2\right)=\frac{d}{dx}\left(x^4-5x^2\right)\\ \frac{d}{dx}\left(y^4\right)-4\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(x^4\right)-5\frac{d}{dx}\left(x^2\right)\\ \frac{d}{dx}\left(y^4\right)-4\frac{d}{dx}\left(y^2\right)=4x^2-10x\end{array}$

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

$\begin{array}{l}\left[\frac{d}{dy}\left(y^4\right)\frac{dy}{dx}\right]-4\left[\frac{d}{dy}\left(y^2\right)\frac{dy}{dx}\right]=4x^2-10x\\ 4y^3\frac{dy}{dx}-4\left[2y\frac{dy}{dx}\right]=4x^2-10x\\ 4y^3\frac{dy}{dx}-8y\frac{dy}{dx}=4x^2-10x\end{array}$

Substitute $\left(0,-2\right)$ for $\left(x,y\right)$,

$\begin{array}{l}4{\left(-2\right)}^3\frac{dy}{dx}-8\left(-2\right)\frac{dy}{dx}=4{\left(0\right)}^2-10\left(0\right)\\ \left(4\right)8\frac{dy}{dx}+16\frac{dy}{dx}=0\\ \frac{dy}{dx}=0\end{array}$

Thus, the slope of the tangent at $\left(0,-2\right)$ is $m=0$.

Substitute $\left(0,-2\right)$ for $\left(x_1,y_1\right)$ and $m=0$ in equation (1),

$\begin{array}{l}y+2=0\left(x-0\right)\\ y+2=0\\ y=-2\end{array}$

Therefore, the equation of the tangent line to the equation $y^2\left(y^2-4\right)=x^2\left(x^2-5\right)$ at $\left(0,-2\right)$ 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 $y^2\left(y^2-4\right)=x^2\left(x^2-5\right)$ at the point $\left(0,-2\right)$.