# How do you convert #y=x-2y+x^2-3y^2 # into a polar equation?

for polar conversion

for

substituting from above.

#cancel(r)sintheta=cancel(r)costheta- 2cancel(r)sintheta+cancel(r^2)^rcos^2theta-3cancel(r^2)^rsin^2theta#

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Making substitutions and reorganizing for the explicit form

This represents a hyperbola, with center at (-1/2, -1/2). See graph.

graph{x^2-3y^2+x-3y=0 [-2.5, 2.5, -1.25, 1.25]}

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To convert the equation ( y = x - 2y + x^2 - 3y^2 ) into a polar equation, you'll need to express ( x ) and ( y ) in terms of ( r ) and ( \theta ), where ( r ) represents the distance from the origin and ( \theta ) represents the angle from the positive x-axis.

Given that ( x = r \cos(\theta) ) and ( y = r \sin(\theta) ), substitute these expressions into the given Cartesian equation:

( y = x - 2y + x^2 - 3y^2 )

( r \sin(\theta) = r \cos(\theta) - 2(r \sin(\theta)) + (r \cos(\theta))^2 - 3(r \sin(\theta))^2 )

Now, simplify and solve for ( r ) in terms of ( \theta ) to obtain the polar equation.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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