Finding points of horizontal and vertical tangency

[rafeeki92]

Ok, so given the graph:
4x2 + y2 - 8x + 4y + 4 = 0

Find the coordinates of the points of horizontal and vertical tangents

So for the horizontal tangents I did:
8x + 2y(dy/dx) - 8 + 4(dy/dx) = 0
(dy/dx)(2y+4) = -8x - 8
dy/dx = (-8x-8)/(2y+4)

So to find the x-coordinates:
(-8x-8)/(2y+4) = 0
-8x-8 = 0
x = -1


For the vertical tangents:
8x(dx/dy) + 2y - 8(dx/dy) + 4 = 0
(dx/dy)(-8x - 8) = 2y+4
dx/dy = (2y+4)/(-8x-8)

To find the y-coordinates:
2y+4 = 0
y = -2


My question is...how can I find the y-coordinates of the points of tangency for the horizontal tangents, and how can I find the x-coordinates of the points of tangency for the vertical tangents?

 

[Deleted member 4993]

Ok, so given the graph:
4x2 + y2 - 8x + 4y + 4 = 0

Find the coordinates of the points of horizontal and vertical tangents

So for the horizontal tangents I did:
8x + 2y(dy/dx) - 8 + 4(dy/dx) = 0
(dy/dx)(2y+4) = -8x + 8 <<< This is not correct
dy/dx = (-8x-8)/(2y+4)

So to find the x-coordinates:
(-8x-8)/(2y+4) = 0
-8x-8 = 0
x = -1

4x[sup:1zahci7k]2[/sup:1zahci7k] + y[sup:1zahci7k]2[/sup:1zahci7k] - 8x + 4y + 4 = 0

4(-1)[sup:1zahci7k]2[/sup:1zahci7k] + y[sup:1zahci7k]2[/sup:1zahci7k] - 8(-1) + 4y + 4 = 0

4 + y[sup:1zahci7k]2[/sup:1zahci7k] + 8 + 4y + 4 = 0

y[sup:1zahci7k]2[/sup:1zahci7k] + 4y + 16 = 0

and continue....




For the vertical tangents:
8x(dx/dy) + 2y - 8(dx/dy) + 4 = 0
(dx/dy)(-8x - 8) = 2y+4
dx/dy = (2y+4)/(-8x-8)

To find the y-coordinates:
2y+4 = 0
y = -2


My question is...how can I find the y-coordinates of the points of tangency for the horizontal tangents, and how can I find the x-coordinates of the points of tangency for the vertical tangents?

 

[BigGlenntheHeavy]

\(\displaystyle One \ way:\)

\(\displaystyle 4x^{2}-8x+y^{2}+4y+4=0\)

\(\displaystyle 4(x^{2}-2x)+y^{2}+4y=-4\)

\(\displaystyle 4(x-1)^{2}+(y+2)^{2}=-4+4+4\)

\(\displaystyle 4(x-1)^{2}+(y+2)^{2}=4\)

\(\displaystyle \frac{(x-1)^{2}}{1^{2}}+\frac{(y+2)^{2}}{2^{2}}=1\)

\(\displaystyle Hence, \ we \ have \ an \ ellipse \ with \ H.T. \ at \ (1,0),(1,-4) \ and \ V.T. \ at \ (0,-2),(2,-2).\)

\(\displaystyle See \ graph.\)

[attachment=0:kc4qfe0t]all.jpg[/attachment:kc4qfe0t]