Find the points at which the graph of 2x^2+2y^2- 20x +12y +3 = 0 has a vertical and horizontal tangent line.

Compute the derivative dy/dx and check where it is zero (for horizontal tangents) or undefined (for vertical tangents).

2x² + 2y² - 20x + 12y + 3 = 0

d/dx [2x² + 2y² - 20x + 12y + 3] = 0

4x + 4y dy/dx - 20 + 12 dy/dx = 0

(4y + 12) dy/dx = 20 - 4x

(y + 3) dy/dx = 5 - x

dy/dx = (5 - x) / (y + 3)

• Horizontal tangents:

dy/dx = 0   →   5 - x = 0   →   x = 5

Solve for y when x = 5 :

2•5² + 2y² - 20•5 + 12y + 3 = 0

2y² + 12y - 47 = 0

y = (-6 ± √(130))/2

So there are two horizontal tangents at the points

(5, (-6 - √(130))/2) and (5, (-6 + √(130))/2)

• Vertical tangents:

1/(dy/dx) = 0   →   y + 3 = 0   →   y = -3

Solve for x when y = -3 :

2x² + 2•(-3)² - 20x + 12•(-3) + 3 = 0

2x² - 20x - 15 = 0

x = (10 ± √(130))/2

So there are two vertical tangents at the points

((10 - √(130))/2, -3) and ((10 + √(130))/2, -3)

Alternatively, you can complete the square to identify the equation of a circle:

2x² + 2y² - 20x + 12y + 3 = 0

2 (x² - 10x) + 2 (y² + 6y) = -3

2 (x² - 10x + 25 - 25) + 2 (y² + 6y + 9 - 9) = -3

2 (x - 5)² - 50 + 2 (y + 3)² - 18 = -3

2 (x - 5)² + 2 (y + 3)² = 65

(x - 5)² + (y + 3)² = 65/2

which is a circle centered at (5, -3) with radius √(65/2). The horizontal tangents occur at the points where the x term vanishes (x = 5), and the vertical ones where y vanishes (y = -3).