
Yefim S. answered 10/13/19
Math Tutor with Experience
We differentiate left and rite sides by x using implicit differentiation: 128x + 50y•y' + 640 - 400y' = 0.
Now we solve this equation for y': 128x + 640 = (400 - 50y)y'.
from here y' = (128x + 640)/(400 - 50y).
If tangent line vertical slope or y' DNE, so 400 - 50 y = 0, y = 400/50 = 8.
Now wde put y = 8 in given equation to get x: 64x2 + 25 • 82 + 640x - 400•8 + 1600 = 0,
64x2 + 640x = 0, or x2 + 10x = 0, x(x+ 10) = 0. x = 0 or x = -10. We have vertical tangent lines at 2 points:
(0,8) and (-10, 8).
If tangent line horizontal then slope or y' = 0. From here 128x + 640 = 0, x = -640/128 = -5. Now we put x = -5 in original equation:64(-5)2+ 25y2+ 640(-5) - 400y + 1600 = 0, 25y2 - 400y = 0, y = 0 or y = 16. We have two points with horizontal tangent line: (-5, 0) and (-5, 16)