# Convert the polar equation into rectangular coordinates

## Homework Statement

r^2= 2cos^2 θ+3sin^2θ

X= r cosθ
y= r sinθ

## The Attempt at a Solution

√r=√2cos^2 θ+3sin^2θ
r = 2 cos θ+ 3 sin θ
r = 2x + 2 y.
I doubt that i even got close to the correct answer so I like to ask anyone who knows how to deal with this type of equation, Did I make the right moves ( steps ) or did messed it all up ?and if i did, will you please help ( guide ) me to the right way to correct way of doing these type of equations. thanks in advance.

## The Attempt at a Solution

Mark44
Mentor
$$\sqrt{a^2 + b^2} \neq a + b$$
√r=√2cos^2 θ+3sin^2θ
r = 2 cos θ+ 3 sin θ
r = 2x + 2 y

I see what you did in your first equation, but the second one doesn't follow from the first, for the reason I gave above. The third doesn't follow from the second.

The relationship is x = rcos θ and y = rsin θ. How did you go from the second equation to the third?

Rather than take square root of both sides, start making replacements using the polar equivalents for x and y. Also, there is a useful formula involving x, y, and r.

i know that r =X^2+ Y^2
and what i basically did in the 3rd equation was that i plugged in x instead of Cos θ and Y for Sin θ.

gabbagabbahey
Homework Helper
Gold Member
i know that r =X^2+ Y^2
and what i basically did in the 3rd equation was that i plugged in x instead of Cos θ and Y for Sin θ.

But x isn't Cos θ : it's r*Cos θ or, in terms of x and y $$\sqrt{x^2+y^2}\cos \theta$$ and so $$\cos \theta=\frac{x}{\sqrt{x^2+y^2}}$$ ...how about Sin θ?

Mark44
Mentor
i know that r =X^2+ Y^2
Then you know something that isn't true, so isn't really worth knowing.

Then you know something that isn't true, so isn't really worth knowing.

okay that was kinda harsh! But he's right, you're forgetting something in that r $$\neq$$ x2+y2 what does r equal? Or maybe, what does r2 equal?

HallsofIvy
Homework Helper

## Homework Statement

r^2= 2cos^2 θ+3sin^2θ
Multiplying on both sides by $r^2$ gives $r^4= 2r^2cos^2 \theta+ 3r^2 sin^2 \theta$ or $(r^2)^2= 2(r cos \theta)^2+ 3(r sin \theta)^2$.

X= r cosθ
y= r sinθ

## The Attempt at a Solution

√r=√2cos^2 θ+3sin^2θ
r = 2 cos θ+ 3 sin θ
r = 2x + 2 y.
I doubt that i even got close to the correct answer so I like to ask anyone who knows how to deal with this type of equation, Did I make the right moves ( steps ) or did messed it all up ?and if i did, will you please help ( guide ) me to the right way to correct way of doing these type of equations. thanks in advance.