Lastly, first correct answer gets best answer

Find the points at which the graph of the equation has a vertical or horizontal tangent line. (Order your answers from smallest to largest x, then from smallest to largest y.)

49x^2 +9y^2 -294x +144y+576=0

When x = 3, y1 = - 15, y2 = - 1, points (3, - 15), (3, - 1)

When y = - 8, x1 = 0, x2 = 6, points (0, - 8), (6, - 8)

49x^2 +9y^2 -294x +144y+576=0 ---------------------------- 1
or Differentiating with respect to x,
98x+ 18y(dy/dx) -294 +144(dy/dx) = 0 or
9(dy/dx)(y+8)=-(49x-147) or
dy/dx = -[(49x-147)/{9(y+8)]] --------------------------------- 2

Case 1: Horizontal line.
for horizontal tangent, from 2, dy/dx = 0 or x = 3
Substituting in 1 we get
9y^2 +144y +49*9 -294*3 +576 = 0, dividing by 9 we get
y^2 +16y + 49-98 + 64 = 0 or
y^2 +16y +15 = 0 or
(y+15)(y+1) = 0 or y = -1 and y = -15
So the points of horizontal tangent are (3,-15) and (3, -1)

Case 2: Vertical line.
for vertical tangent dy/dx = ∞, from 2, y = -8
Substituting in 1 we get
9*64 +144(-8)+49x^2 -294x+576 = 0 or
576 -1152 +576 +49x^2 -294x = 0 or
49x^2-294x = 0 or
x^2 -6x = 0 or
x(x-6) = 0or
x = 0 or 6
So the points for vertical tangent line are: (0, -8) and (6, -8)