How do you convert #4=(x4)^2+(y5)^2# into polar form?
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To convert the equation (4 = (x  4)^2 + (y  5)^2) into polar form, we can follow these steps:

Replace (x) and (y) with their respective polar coordinate representations: [ x = r\cos(\theta) \quad \text{and} \quad y = r\sin(\theta) ]

Substitute these expressions into the given equation: [ 4 = (r\cos(\theta)  4)^2 + (r\sin(\theta)  5)^2 ]

Expand and simplify the equation: [ 4 = r^2\cos^2(\theta)  8r\cos(\theta) + 16 + r^2\sin^2(\theta)  10r\sin(\theta) + 25 ]

Combine like terms and rearrange: [ 4 = r^2(\cos^2(\theta) + \sin^2(\theta))  8r\cos(\theta)  10r\sin(\theta) + 41 ]

Recognize that (\cos^2(\theta) + \sin^2(\theta) = 1) (the Pythagorean identity), and substitute: [ 4 = r^2  8r\cos(\theta)  10r\sin(\theta) + 41 ]

Rearrange to isolate (r): [ r^2  8r\cos(\theta)  10r\sin(\theta) = 37 ]

Complete the square for (r): [ r^2  8r\cos(\theta) + 16\cos^2(\theta) + 10r\sin(\theta) + 25\sin^2(\theta) = 37 + 16\cos^2(\theta) + 25\sin^2(\theta) ]

Factor the perfect square trinomials: [ (r  4\cos(\theta))^2 + (5\sin(\theta))^2 = 4 ]

Divide both sides by 4 to isolate (r): [ \frac{(r  4\cos(\theta))^2}{4} + \frac{(5\sin(\theta))^2}{4} = 1 ]

Recognize that (r^2 = x^2 + y^2) and (a^2 = (r  h)^2) and (b^2 = (r  k)^2) (where (h) and (k) are the coordinates of the center of the ellipse) to rewrite the equation: [ \frac{(x  4)^2}{2^2} + \frac{(y  0)^2}{(\frac{5}{2})^2} = 1 ]
Thus, the equation (4 = (x  4)^2 + (y  5)^2) in polar form is (\frac{(x  4)^2}{2^2} + \frac{(y  0)^2}{(\frac{5}{2})^2} = 1).
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When evaluating a onesided 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 onesided 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 onesided 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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