x^2/a^2 + y^2/b^2 = 2

know have to squar them... but really confused....

pleasee help mee...????

## Trigonometry hardest questions

# 2 Answers

Not clear on what you are asking?! The equation fits into the general formula for an ellipse,

(x-h)^{2}/a^{2} + (y-k)^{2}/b^{2 }= 1

The length of the "major" (longer) axis of symmetry is "a"; the length of the "minor" (shorter) axis is "b"; there are two vertices and two co-vertices. You can find all the details at

http://www.clausentech.com/lchs/dclausen/algebra2/ellipses.htm.

You can fit your formula into the general equation by dividing both sides by 2.

Square each to get x^{2}/a^{2} cos^{2} Θ + xy/ab cos Θ sin Θ + y^{2}/b^{2} sin ^{2} Θ = 1 and

x^{2}/a^{2} sin^{2} Θ - xy/ab cos Θ sin Θ + y^{2}/b^{2} cos ^{2} Θ = 1. Now add them together - note that

1 - your center terms drop out

2 - factor common terms and you will be left with

x^{2}/a^{2} (cos^{2} Θ + sin ^{2}Θ) + y^{2}/b^{2} (cos^{2} Θ + sin
^{2}Θ) = 2

3 - What does (cos^{2} Θ + sin ^{2}Θ) equal? Replace it and solve.

Good luck.

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