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Jorge
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#1
Report Thread starter 14 years ago
#1
I've gone wrong somewhere and can't see where.


x= 3cosA , y = 2 sinA

---

x^2 = 9cos^2A
x^2 = 9(1 - sin^2A)
x^2 = 9 - 9sin^2A
sin^2A = (9 - x^2)/9



y^2 = 4 sin^2A

y^2 = 4[(9 - x^2)/9]

y^2 = (36 - 4x^2)/9

9y^2 = 36 - 4x^2

4x^2 + 9y^2 = 36

---

However the book's answer is

x^2/9 + y^2/4 = 1

Where did I go wrong?

Thanks

jorge
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BCHL85
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#2
Report 14 years ago
#2
Nothing is wrong, just devide both sides by 36.
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