Excellent work. The equation x^2 + y^2 = 4 x y is indeed correct. It was very clever to multiply both sides by r in order to get r cos(theta) on the right-hand side.
To graph this equation we can proceed as follows, completing the square on x and solving for y in terms of x:
x^2-4xy+y^2=0. Complete the square on x^4 - 4 xy by adding and subtracting 4 y^2:
x^2-4xy+4y^2 -4y^2+y^2=0. x^2 - 4 x y + 4 y^2 is (x - 2y)^2. Rearranging we get:
(x-2y)^2=3y^2
x-2y=+-sqrt(3)y
x=(2+-sqrt(3))y
y = 1 / (2 +- sqrt(3) ) x = (2 -+ sqrt(3) ) x
The latter is just a pair of straight lines with slopes 2 - sqrt(3) and 2 + sqrt(3).
You could alternatively have divided both sides of the original equation by r, obtaining sec(theta) = 4 sin(theta). This rearranges to
sin(theta) * cos(theta) = 1/4.
Using sin(theta) = y / sqrt(x^2+y^2) and cos(theta) = x / sqrt(x^2 + y^2) we pretty quickly get back to the same equation as above.
Another way to solve sin(theta) cos(theta) = 1/4 (I'm putting it in smaller type so you can more easily ignore it, which I recommend):
This is an equation in theta only. It is independent of r. So any theta that satisfies this condition will satisfy it for any r, giving us a straight line at angle theta.
If we can find the points on the unit circle that satisfy this equation, we will know which straight lines comprise solutions.
Now on the unit circle, sin(theta) and cos(theta) are given respectively by y and x; since we are on the unit circle we know that y = sqrt(1-x^2).
So the equation sin(theta) cos(theta) = 1/4, on the unit circle, becomes x y = 1/4 subject to the constraint y = sqrt(1-x^2).
The equation therefore becomes
x sqrt(1-x^2) = 1/4, or squaring both sides
x^2 ( 1 - x^2) = 1/16.
This is the same as the quadratic equation u^2 - u + 1/16 = 0, where u = x^2. Solutions are u = (1 +- sqrt( 1 - 1/4) ) / 2 = (1 +- sqrt(3) /2 ) / 2 = 1/2 +- sqrt(3) / 4.
It gets even messier: Remember that u^2 = x. So if we plug this expression in for x in y = sqrt (1 - x^2), then divide this expression for y by the square root of the expression for u, we get the slopes 2 +- sqrt(3).
Sir,
I was going over some problems and thought this one was a beaut, if I could only solve it.
Transform the equation: r sec( Theta)=4 r sin (theta) to rectangular Coordinates and graph.
Answer:
r sec( Theta)=4 r sin (theta)
r*(1/cos(theta)= 4y
r=4y*cos(theta)
r* (r=4y*cos(theta)
r^2=r*(4y*cos(theta)
x^2 +y^2= 4y*r cos (theta)
x^2 +y^2= 4y *x
At this point I get stuck although when I look at ....r sec( Theta)=4 r sin (theta) I can see r= 4 r* sin (theta) * cos (theta) and the r/2 = 2r*(sin (theta) * cos (theta) looks like a double angle but this still doesn't help me get to the coordinates.
What's your thoughts..