Find the orthogonal trajectory of the circle x^2+y^2+2fy +1=0
Hmmm. The orthognal curves to a circle are its radii.
So, since your circle is
x^2+(y+f)^2 = r^2
you want all the lines through (0,f)
That would be y=kx+f
To find the orthogonal trajectory of a given curve, follow these steps:
Step 1: Write the equation of the given curve in the form f(x, y) = 0.
The equation of the circle x^2 + y^2 + 2fy + 1 = 0 can be rearranged as:
x^2 + y^2 + 2fy + 1 = 0
x^2 + y^2 + 2fy = -1
2fy = -x^2 - y^2 - 1
fy = (-1/2)x^2 - (1/2)y^2 - (1/2)
Step 2: Differentiate the function fy(x, y) with respect to y.
Differentiating both sides of fy = (-1/2)x^2 - (1/2)y^2 - (1/2) with respect to y, we get:
df/dy = -y
Step 3: Find the negative reciprocal of df/dy to get the slope of the orthogonal trajectory.
The negative reciprocal of -y is 1/y.
Step 4: Write the differential equation for the orthogonal trajectory using the slope from step 3.
Since the slope of the orthogonal trajectory is 1/y, the differential equation for the orthogonal trajectory is:
dy/dx = y
Step 5: Solve the differential equation from step 4 to find the equation of the orthogonal trajectory.
The differential equation dy/dx = y is a separable equation, which can be solved by separating the variables:
(dy/y) = dx
Integrating both sides, we get:
ln|y| = x + C
where C is the constant of integration.
Taking the exponential of both sides, we have:
|y| = e^(x+C)
Since the absolute value of y can be positive or negative, we can split it into two cases:
1. y = e^(x+C) for y > 0
2. y = -e^(x+C) for y < 0
Therefore, the equation of the orthogonal trajectory is given by:
y = Ce^x, where C is a constant.