Finding the equation of the line tangent to $y^2(y^2-4)=x^2(x^2-5)$ I am looking to find the equation of the line tangent to 
$$y^2(y^2-4)=x^2(x^2-5)$$ 
at the point $(0,-2)$.
I have a feeling I need to implicitly differentiate here? 
Am I on the right track?
What do I do after finding $y'$ to actually find the solution?  Like what steps do I take to find the tangent line?
Not asking for the solution but a push in the right direction would be helpful, although a solution would be nice to look over. 
 A: HINT
Yes, you need to implicitly differentiate. In order to describe a line, you need a point and a slope. You already have the point, so you just need to extract the slope from the implicit differentiation.
A: Yes you do need to use implicit differentiation. When finding a tangent line, you nearly always need the point-slope formula:
$$y_2 - y_1 = m(x_2 - x_1)$$
Solution to the implicit differentiation is below (to check your work). Simply hover your mouse over the grey box.

 $y^2(y^2 - 4) = x^2(x^2 -5) \\\\$ Multiplying the polynomials gets us to
 $y^4 - 4y^2 = x^4 - 5x^2$. Taking the derivative with respect to $x$ gets us: $4y^3y' - >! 8yy'=4x^3 - 10x$. Factoring to get $y'$ by itself: $y'(4y^3 - 8y) = 4x^3 - 10)$. Divide through to get $y'$ by itself: $y' = \dfrac{4x^3 - 10x}{4y^3-8y}$. You could make your life a bit easier by factoring this into $y' = \dfrac{2x(2x^2 - 5)}{4y(y^2-2)}$. You could cancel out a factor of $2$ to get $y' = \dfrac{x(2x^2 - 5)}{2y(y^2-2)}$. To find the slope, plug in your points $x = 0, y = -2$ into our equation for $y'$ to find the slope of the line. Note that the slope is $0$. To find the equation of the tangent line, use that value for $m$ you just found ($m=0$) and your given points into the point-slope formula and you find that the tangent line is $y=-2$.

Added:
Our expression for $y'$ is:
$$y' = \dfrac{2x(2x^2 - 5)}{4y(y^2-2)}$$
We were given a point $(0,-2)$. So, $x = 0, y = -2.$ Plugging this into the expression above yields:
$$y' = 0.$$
So, our slope of the line tangent to the curve $y^4-4y^2 = x^4-5x^2$ is zero. Now, using the point-slope formula, we have:
$$y - -2 = 0(x-0)$$
$$y+2 = 0 \implies y = -2$$
So, the tangent line is simply the horizontal line $y=-2.$
A: Differentiate both sides with respect to $x$. On the right, we get $4x^3-10x$.  On the left we get $(4y^3-8y)\frac{dy}{dx}$. 
Now (crucial) we need to evaluate the derivative at $(0,-2)$.  At this point, $4x^3-10x$ is $0$, and $4y^3-8y$ is not. So $\left(\frac{dy}{dx}\right)_{(0,-2)}=0$.  
Slope is $0$, we pass through $(0,-2)$, so the tangent line is horizontal, and has equation $y=-2$.
