find the equations of ALL horizontal tangents to the curve y^2=(x^2+4)/x, if any exist
This is what I have so far:
xy^2= x^2+4
2xyDy/Dx =2x-y^2
dy/dx = (2x- y^2)/2xy
(2x- y^2)/2xy = 0
2x-y^2
I don't know what to do next
So far, so good. We know that for a horizontal tangent, y'=0, so
2x-y^2 = 0
But, we already know that
y^2 = (x^2+4)/x, so
2x - (x^2+4)/x = 0
2x^2 - x^2-4 = 0
x^2 = 4
x = 2 or -2
At x = 2, y = 4
At x = -2, y = -4
Those are the horizontal tangents
To find the equations of all horizontal tangents to the curve, we need to find the points where the derivative dy/dx is equal to zero.
From your work so far, you have the derivative expression:
dy/dx = (2x - y^2) / (2xy)
To find the horizontal tangents, we set dy/dx equal to zero:
(2x - y^2) / (2xy) = 0
Since the numerator of the fraction is set equal to zero, we can set each factor equal to zero:
2x - y^2 = 0
Now, we have one equation with two variables, so we need another equation to solve for both x and y simultaneously. The other equation we can use is the original equation of the curve:
y^2 = (x^2 + 4) / x
We can substitute this equation into the previous one to eliminate y:
2x - [(x^2 + 4) / x]^2 = 0
Now, we simplify this equation by multiplying both sides by x^2, which gives:
2x^3 - (x^2 + 4)^2 = 0
This equation is a cubic equation in terms of x, and it may be difficult to solve exactly. However, we can still find the coordinates of the points where dy/dx is equal to zero by using numerical methods or graphing software.
Once you have the x-coordinates of these points, you can substitute them back into the original equation y^2 = (x^2 + 4) / x to find the corresponding y-coordinates. Then, you can write the equations of the horizontal tangents using the point-slope form of a line:
y - y-coordinate = 0(x - x-coordinate)
Simplifying this equation will give you the equations of all the horizontal tangents to the curve.