Find the points at which the graph of the equation has a vertical or horizontal line:
25x^2 + 16y^2 + 200x -160y + 400 = 0
Need working out too please.
2007-10-29 10:08:05 · 4 answers · asked by Raja A 1 in Science & Mathematics ➔ Mathematics
25x² + 16y² + 200x -160y + 400 = 0
∂/∂x 25x² + 16y² + 200x -160y + 400 = ∂/∂x 0
50x + 32yy' + 200 -160y' = 0
50x + 200 = 160y' - 32yy'
50x + 200 = [ 160 - 32y ] y'
[ 50x + 200 ] / [ 160 - 32y ] = y'
[ 25x + 100 ] / [ 80 - 16y ] = y'
2007-10-29 10:12:47 · answer #1 · answered by Anonymous · 0⤊ 0⤋
25x^2 + 200x + 400= 160y - 16y^2
(50x+200) dx= (160-32y)dy
dy/dx = (160-32y)/(50x+200)
2007-10-29 10:16:52 · answer #2 · answered by Anonymous · 0⤊ 1⤋
take d/dx
50*x + 32*y*dy/dx + 200 - 160*dy/dx = 0
dy/dx*(32y - 160) = -(50x + 200)
dy/dx = -50/32*(x + 4)/(y - 5)
dy/dx = 0 => x = -4 (as long as y neq 5)
dy/dx = inf => y = 5 (as long as x neq -4)
Then solve for the missing y or x depending on which case you're doing. This is an ellipse so you should get two points for each.
2007-10-29 10:26:43 · answer #3 · answered by Anonymous · 0⤊ 0⤋
50x+32y*y´+200-160y´=0
y´= (-50x-200)/(32y-160)
Horizontal line at x= -4
vertical line y=5
2007-10-29 10:36:39 · answer #4 · answered by santmann2002 7 · 0⤊ 0⤋