at what points on the hyperbola xy=12 is the tangent line parrallel to the line y= -3x
To find the points on the hyperbola xy = 12 where the tangent line is parallel to the line y = -3x, we need to first take the derivative of the hyperbola equation.
The derivative of xy = 12 can be found using implicit differentiation. We differentiate both sides of the equation with respect to x:
d/dx(xy) = d/dx(12)
Using the product rule, we get:
y * dx/dx + x * dy/dx = 0
Simplifying this equation, we have:
1 * y + x * dy/dx = 0
Rearranging the terms, we have:
dy/dx = -y/x
Now we have the derivative of the hyperbola equation.
Since the derivative represents the slope of the tangent line, we want to find where this derivative is equal to the slope of the line y = -3x.
The slope of y = -3x is -3.
Therefore, we set dy/dx = -3:
-y/x = -3
To solve this equation for y, we can multiply both sides by x:
-y = -3x
Now, divide both sides by -1:
y = 3x
So, the slope of the tangent line is 3.
Now that we have the slope of the tangent line, we can substitute it back into the original hyperbola equation to find the points where the tangent line is parallel to y = -3x.
xy = 12
Substituting y = 3x:
x * 3x = 12
3x^2 = 12
Dividing both sides by 3:
x^2 = 4
Taking the square root of both sides:
x = ±2
Therefore, the x-values of the points where the tangent line is parallel to y = -3x are x = 2 and x = -2.
Now, substituting these x-values back into the original hyperbola equation, we can find the corresponding y-values:
For x = 2:
y = 12/2 = 6
So, one point is (2, 6).
For x = -2:
y = 12/-2 = -6
Another point is (-2, -6).
Thus, the two points on the hyperbola xy = 12 where the tangent line is parallel to the line y = -3x are (2, 6) and (-2, -6).
slope= -3
y=12/x
y'=-3=-12/x^2
x^2=4
x=+-2
check that