Question

Given the equation of a closed curve is x^2-2x+4y^2+16y+1=0.Find the coordinates of the two points where the lines tangent to the curve are vertical.

Answer 1

2x - 2 + 8y dy/dx + 16dy/dx = 0
dy/dx(8y + 16) = 2 - 2x
dy/dx = (2-2x)/(8y + 16)

if the tangents are vertical, then their slopes are undefined, that is, the denominator of dy/dx = 0
8y + 16 = 0
y = -2
sub into original equation:
x^2 - 2x + 8 - 32 + 1 = 0
x^2 - 2x - 23 = 0

x = (2 ± √96)/2
= 1 ± 2√6

the points are (1+2√6 , -2) and (1 - 2√6 , -2)

Answer 2

This can be easily confirmed by noting that the equation is an ellipse:

(x-1)^2/16 + (y+2)^2/4 = 1

The major axis is on the line y = -2, so that's where the tangents are vertical.