Show that r = 1/4 - 3Cos¦È is polar form of an ellipse and find its centre.
pls can anyone could help with this.
To show that the equation r = 1/4 - 3Cosθ represents the polar form of an ellipse, we need to simplify the equation and observe its properties.
First, let's manipulate the equation to make it more recognizable. We can multiply through by 4 to eliminate the denominator:
4r = 1 - 12Cosθ
Next, we can square both sides to eliminate the cosine term:
(4r)^2 = (1 - 12Cosθ)^2
16r^2 = (1 - 12Cosθ)^2
Now, let's expand the right side of the equation:
16r^2 = 1 - 24Cosθ + 144Cos^2θ
Simplifying further, we have:
16r^2 = 1 - 24Cosθ + 144(1 - Sin^2θ)
16r^2 = 1 - 24Cosθ + 144 - 144Sin^2θ
16r^2 = 145 - 24Cosθ - 144Sin^2θ
Now, we can rewrite the equation in terms of r^2:
r^2 = (145/16) - (24/16)Cosθ - (144/16)Sin^2θ
Simplifying:
r^2 = (145/16) - (3/2)Cosθ - 9Sin^2θ
We can see that the equation is in the form:
r^2 = A - BCosθ - DSin^2θ
This represents the polar equation of an ellipse, where the center of the ellipse is given by (BCosθ, DSinθ), or in this case, (-B/2, -D/2).
Comparing with the given equation:
r^2 = (145/16) - (3/2)Cosθ - 9Sin^2θ
We can conclude that it represents the polar form of an ellipse, with the center of the ellipse at (-(-3/2)/2, -(-9)/2), which simplifies to (3/4, 9/2).