i want to know how the solution comes for the following:
if x cosA/a + y sinA/b=1, and x sinA/a - y cosA/b=1, prove that x square/a square+ysquare/bsquare=2. i hope you understood what "square" meant. if you didn't, then here's the explanation :2 square= 4. i didn't know to insert the exponent. so please adjust as the question is correct.
2006-06-17 04:43:33 · 7 answers · asked by janani 2 in Science & Mathematics ➔ Mathematics
check the statement of the problem
and 3squared=3^2=9
2006-06-17 05:05:41 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
Given xcosA/a + ysinA/a = 1 and x sinA/a - y cosA/b=1
Squaring and adding both sides of each equation, we get
x^2 (cosA)^2/a^2 + y^2 (sinA)^2/a^2 - 2xysinA cosA/ab +
x^2 (sinA)^2/a^2 + y^2 (cosA)^2/a^2 - 2xysinA cosA/ab = 2
Now the cross product term gets cancelled, leaving
x^2/a^2 + y^2/a^2 = 2 {Since (sinA)^2 + (CosA)^2 = 1}
2006-06-17 05:07:53 · answer #2 · answered by raobn 2 · 0⤊ 0⤋
are you working with parametric equations?
or transformations between rcosA = x ?
Raobn: you have an error with your a 's and b's according to what is given in the statement.
so maybe the question is copied wrong?
2006-06-17 06:02:45 · answer #3 · answered by Mary B 1 · 0⤊ 0⤋
xcosA/a+ysinB/b=1
squaring on both sides
xsquare cossquare/asquare+ysquare sinsquare/asquare+2.x.y.cosA.sinA/ab=1 -------I
xsinA/b-ycosA/b=1
squaring on both sides
x^2sin^2A/b^2+y^2cos^2A/b^2-2xycosAsinA/ab=1 ---------II
add I and II
[x/a]^2+[y/b]^2=1+1=2
since sin^2A+cos^2A=1
hence proved
2006-06-17 06:56:11 · answer #4 · answered by Anonymous · 0⤊ 0⤋
if you are having problems go to www.math.com There is help for anything math on that site.
2006-06-17 04:46:55 · answer #5 · answered by houstonmom77064 3 · 0⤊ 0⤋
square both the equations and add them up afterwards ... u wud get ur solution...
2006-06-17 04:58:54 · answer #6 · answered by Anonymous · 0⤊ 0⤋
I am sure that I don't know .
and also please don't ask your hw problems.
2006-06-17 06:13:39 · answer #7 · answered by ALex 1 · 0⤊ 0⤋