How do I convert the following polar expression to rectangular form?
3 sec (beta)
r= ----------------
4 sec (beta) + 4
[If not clear, the equation is r equals 3 times secant of angle beta divided by 4 times the sec of angle beta plus 4.]
2 answers
wow... this is in algebra? how come i nvr got i problem like this? and i was just in it last year!.....sorry, i dnt know...i dnt knw why i posted this as an answer tho XP....
draw yourself a right-angled triangle with sides x,y, and r so that
cosß = x/r, then secß = r/x
so your equation becomes
r = (3r/x)/(4r/x + 4)
cross-multiply
4r^2/x + 4r = 3r/x
divide by r
4r/x + 4 = 3/x
multiply by x
4r + 4x = 3
4r = 3-4x
square both sides
16r^2 = 9 - 24x + 16x^2
but r^2 = x^2 + y^2
16(x^2+y^2) = 9 - 24x + 16x^2
16x^2 + 16y^2 = 9 - 24x + 16x^2
16y^2 = 9 - 24x
(looks like a horizonatal parabola)
cosß = x/r, then secß = r/x
so your equation becomes
r = (3r/x)/(4r/x + 4)
cross-multiply
4r^2/x + 4r = 3r/x
divide by r
4r/x + 4 = 3/x
multiply by x
4r + 4x = 3
4r = 3-4x
square both sides
16r^2 = 9 - 24x + 16x^2
but r^2 = x^2 + y^2
16(x^2+y^2) = 9 - 24x + 16x^2
16x^2 + 16y^2 = 9 - 24x + 16x^2
16y^2 = 9 - 24x
(looks like a horizonatal parabola)
1 answer