Math253 conversion of ellipse equation PDF

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Math 253 – conversion of ellipse equation to standard and polar forms We derived the equation of an ellipse with center at the origin and both foci on the x-axis as 2

2

x y + 2 =1 2 a b

with a  b . (When b  a , the equation describes an ellipse with foci on the y-axis.) The foci are located at (c, 0) and ( −c, 0) with c = a2 − b2 . If the foci are located on some other horizontal line and the center of the ellipse is at the point ( h, k ) instead of (0, 0) , the equation becomes 2

2

( x − h) ( y − k) + =1 2 2 a b

We are going to start with the equation 16 x 2 − 96 x + 25 y 2 = 256 and put it in the above form. Note that B2 − 4 AC = 0 2 − 4(16)(25) = −1600  0 , so this equation does represent an ellipse, not a circle since A  C . Now we transform it by completing the square: 16( x 2 − 6 x) + 25 y 2 = 256 16( x2 − 6 x + 9 − 9) + 25 y2 = 256

16( x 2 − 6 x + 9) −144 + 25 y 2 = 256 16( x − 3)2 + 25 y2 = 400

Dividing both sides by 400 and simplifying puts this equation in the form 2

2

( x − 3) y + =1 25 16

So a = 5 , b = 4 , and the center of the ellipse is at (3, 0) . Since c = 52 − 42 = 3 , the foci are 3 units to the left and right of the center point which puts the foci at (0, 0) and (6, 0) . Because one of the foci is at the origin, we can also convert this equation to polar form. We return to the original equation and substitute to convert it to polar coordinates: 16 x 2 − 96 x + 25 y 2 = 256 16( r cos  ) 2 − 96( r cos  ) + 25( r sin  ) 2 = 256 (16 cos2  + 25sin2  ) r2 − (96 cos ) r − 256 = 0

We can use the quadratic formula to solve this equation for r in terms of  . We first use the Pythagorean identity sin 2  = 1 − cos2 to rewrite it as (25 − 9 cos 2  ) r 2 − (96cos ) r − 256 = 0

then identify a, b, and c and substitute into the quadratic formula to obtain r=

+96 cos  962 cos2  + 4(25 − 9 cos 2  )256 2(25 − 9 cos 2  )

The cos2  terms inside the square-root cancel, giving us r=

96cos   25600 96cos  160 = 2 2(25 − 9cos  ) 2(25 − 9cos2  )

We must take the plus sign if we want r to be positive, although the minus sign also gives us a valid equation in polar coordinates for the same ellipse. We factor the denominator as a difference of squares and simplify r as follows: r=

16 96cos  + 160 32(3cos + 5) 16 5 = = = 2(25 − 9cos2  ) 2(5+ 3cos )(5− 3cos ) 5− 3cos 1− 35 cos

Comparing this to the equation on page 5 of section 9.6, we see that this is the equation of an ellipse with eccentricity e = 3 / 5 and one focus at the origin. Note that 0  e  1 ....