Find the point(s) on the graph y=1/x where the graph is parallel to the line 4x+9y=3?

Question

I dont know where to go from here. All I know is that both lines will have teh same slope. I tried to graph it but it didnt help. Any help will be appreciated

Answer 1

y= 1/x
y' = -1/x^2 = -4/9
x = ±3/2, y = ±2/3

Answer 2

1) for clarity, convert the second equation into function, or "slope-intercept" form:

4x + 9y = 3

y2 = -(4/9)x + 1/3

2) find the derivatives:

y1 = 1 / x
dy1/dx = -1 / x²

y2 = -(4/9)x + 1/3
dy2/dx = -4/9

3) set dy1/dx = dy2/dx

-1 / x² = - 4/9

4) solve for x !

Hope that make sense......
~W.O.M.B.A.T.

Answer 3

I have them intersecting at X=3/13 and y= 13/3

Y=1/x

4x+9y=3
therfore:
4x+9x=3
therefore:
13x=3

x=3/13

y=13/3

Parallel lines don't intersect.

Answer 4

dy = ln(x). This is essentially the slope at x

4x + 9y = 3
slope = -4/9

ln(x) = -4/9
x = exp(-4/9) <- natural exponent e

Answer 5

Umm... are you sure you wrote that down correctly? y=1/x is an inverse variation function.