Finding points with vertical tangents

Consider the closed curve in the xy plane given by this expression. Here, find the coordinates of the two points on the curve where the line tangent to the curve is vertical. So, pause this video and see if you could have a go at it.

I don't know what the exact shape of this closed curve is, but if I were to draw some type of a closed curve, maybe it looks something like this. This isn't the one that's right over here. This one also has two points where my tangent line is vertical. At one point would be right over there; another point would be right over there.

Now, how do we figure this out? Well, what we could do is use implicit differentiation to find the derivative of y with respect to x and think about the x and y values that would give us a situation where that derivative is non-zero in the numerator and zero in the denominator. So let's do that.

Let me rewrite everything I have: (x^2 + 2x + y^4 + 4y = 5). I want to take the derivative with respect to x of both sides of this equation. I'm trying to find an expression for the derivative of y with respect to x. So what am I going to get? This is going to be equal to (2x + 2 +) the derivative of this with respect to y is (4y^3) and then times the derivative of y with respect to x; that's just straight out of the chain rule. Plus, the derivative of this with respect to y is (4) times the derivative of y with respect to x—once again straight out of the chain rule—is equal to, whoops, I want to take the derivative with respect to x here, is equal to (0).

Now we just have to solve for (\frac{dy}{dx}). A couple of things we could do: we could take the (2x + 2) and subtract it from both sides, and we could also factor out a (4\frac{dy}{dx}) out of this stuff right over here. So let's do that: let's subtract the (2x + 2) from both sides and factor out the (4\frac{dy}{dx}).

We will get (4 \cdot \frac{dy}{dx} \cdot (y^3 + 1) = -2(x + 1)). Now I just have to divide both sides by (4(y^3 + 1)), and I'm going to get the derivative of y with respect to x is equal to (\frac{-2(x + 1)}{4(y^3 + 1)}). Actually, this can be rewritten as being equal to (-\frac{x + 1}{2(y^3 + 1)}); I just divided the numerator and the denominator by (2).

Now, why is this useful? Well, we can think about what y-values—because y is the only variable we have in the denominator here—would make the denominator equal (0) and then find the corresponding x-values for those y-values by going to our original equation.

Well, this is going to be (0) when (y = -1). So when (y = -1), let's figure out what x is. To do that we just have to substitute (y = -1) back in our original equation and then solve for x.

Let's do that; let me clear this out since I need that real estate. If we go back and we substitute (y = -1) up here, we're going to get:

[ x^2 + 2x + 1 + 1 - 4 = 5. ]
This is going to be (-3). Subtract (5) from both sides, you get (x^2 + 2x - 8 = 0). This is just simple factoring, so it's going to be ((x + 4)(x - 2) = 0).

What two numbers, when I take the product, I get (-8)? Four and negative two. When I add four and negative two, I get a positive (2); there it is equal to (0). So (x) is equal to (-4) or (x) is equal to (2) when (y) is equal to (-1).

To answer their question, find the coordinates of the two points on the curve where the line tangent to the curve is vertical. Well, the answer here would be—get a little bit of a drum roll—it would be the points ((-4, -1)) and ((2, -1)), and we're done.