How do you find the coordinates of the two points on the #x^2-2x+4y^2+16y+1=0# closed curve where the line tangent to the curve is vertical?

That is the standard equation of an ellipse.

and its semi-axes are:

graph{x^2-2x+4y^2+16y+1=0 [-10, 10, -5, 5]}

The points in which the tangent is vertical are easy to find, also looking the graph:

and

To find the coordinates of the two points on the given closed curve where the tangent line is vertical, we need to determine the points where the derivative of the curve with respect to x is undefined.

First, let's rewrite the equation of the curve in standard form: x^2 - 2x + 4y^2 + 16y + 1 = 0.

Next, we differentiate the equation with respect to x, treating y as a function of x using implicit differentiation.

Taking the derivative of the equation, we get: 2x - 2 + 8yy' + 16y' = 0.

To find the points where the tangent line is vertical, we need to find the values of x that make the derivative undefined. In other words, we need to find the values of x where the denominator of y' becomes zero.

From the equation 2x - 2 + 8yy' + 16y' = 0, we can isolate y' by dividing through by 8y + 16, giving us: y' = (2 - 2x) / (8y + 16).

To make y' undefined, the denominator (8y + 16) must equal zero. Solving this equation, we find y = -2.

Substituting y = -2 back into the original equation of the curve, we get: x^2 - 2x + 4(-2)^2 + 16(-2) + 1 = 0.

Simplifying this equation, we have: x^2 - 2x + 16 = 0.

Solving for x using the quadratic formula, we find two possible values: x = 1 + 3√3 and x = 1 - 3√3.

Substituting these x-values back into the original equation, we can find the corresponding y-values.

Therefore, the coordinates of the two points on the curve where the tangent line is vertical are: (1 + 3√3, -2) and (1 - 3√3, -2).