If the area of the ellipse above can be represented as , where and are coprime positive integers, find .
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Using a x 2 + b x y + c y 2 + d x + e y = 0 and the points I chose above to generate the ellipse.
(1) ( 0 , − 4 ) : 4 c − e = 0 ⟹ c = 4 e
(2) ( 8 , 4 ) : 1 6 a + 8 b + 4 c + 2 d + e = 0
(3) ( 8 , − 4 ) : 1 6 a − 8 b + 4 c + 2 d − e = 0
Subtracting (3) from (2) we obtain: 8 b + e = 0 ⟹ b = − 8 e
(4) ( 5 , 8 ) : 2 5 a + 4 0 b + 6 4 c + 5 d + 8 c = 0
Replacing c = 4 e and b = − 8 e into (2) and (4) ⟹
1 6 a + 2 d = − e
2 5 a + 5 d = − 1 9 e
⟹ a = 1 0 1 1 e and d = − 1 0 9 3 e
⟹ 1 0 1 1 x 2 − 8 1 x y + 4 1 y 2 − 1 0 9 3 x + y = 0 ⟹ 4 4 x 2 − 5 x y + 1 0 y 2 − 3 7 2 x + 4 0 y = 0 ⟹ 1 0 y 2 + 5 ( x − 8 ) y + 4 4 x 2 − 3 7 2 x = 0
Solving for y we obtain:
y = 4 x − 8 ± 2 0 1 3 4 7 5 2 2 0 7 7 4 4 − ( 3 4 7 x − 1 4 4 8 ) 2 .
y ( x ) = 4 x − 8 + 2 0 1 3 4 7 5 2 2 0 7 7 4 4 − ( 3 4 7 x − 1 4 4 8 ) 2 for the portion of the ellipse above the line y = 4 x − 8 .
Setting 2 2 0 7 7 4 4 − ( 3 4 7 x − 1 4 4 8 ) 2 = 0 ⟹ x = 3 4 7 1 4 4 8 ± 4 4 8 1 1 are the points of intersection of the ellipse and the line 4 x − 8 .
Letting a = 3 4 7 1 4 4 8 − 4 4 8 1 1 and b = 3 4 7 1 4 4 8 + 4 4 8 1 1 the area of the ellipse is A = 2 ∫ a b y ( x ) − 4 x − 8 d x =
1 0 1 3 4 7 5 ∫ a b 2 2 0 7 7 4 4 − ( 3 4 7 x − 1 4 4 8 ) 2 d x = 2 5 3 4 7 1 ∫ a b 2 2 0 7 7 4 4 − ( 3 4 7 x − 1 4 4 8 ) 2 d x .
For I ( x ) = ∫ 2 2 0 7 7 4 4 − ( 3 4 7 x − 1 4 4 8 ) 2 d x
Let 3 4 7 x − 1 4 4 8 = 2 2 0 7 7 4 4 sin ( θ ) ⟹ d x = 3 4 7 2 2 0 7 7 4 4 cos ( θ )
⟹ I ( θ ) = 3 4 7 2 2 0 7 7 4 4 ∫ cos 2 ( θ ) d θ = 6 9 4 2 2 0 7 7 4 4 ∫ ( 1 + cos ( 2 θ ) ) d θ = 6 9 4 2 2 0 7 7 4 4 ( θ + sin ( θ ) cos ( θ ) )
⟹ I ( x ) = 6 9 4 2 2 0 7 7 4 4 ( arcsin ( 2 2 0 7 7 4 4 3 4 7 x − 1 4 4 8 ) + 2 2 0 7 7 4 4 3 4 7 x − 1 4 4 8 2 2 0 7 7 4 4 − ( 3 4 7 x − 1 4 4 8 ) 2 ) ⟹
A = ( 3 4 7 ) 2 3 5 5 5 1 9 3 6 ( arcsin ( 2 2 0 7 7 4 4 3 4 7 x − 1 4 4 8 ) + 2 2 0 7 7 4 4 3 4 7 x − 1 4 4 8 2 2 0 7 7 4 4 − ( 3 4 7 x − 1 4 4 8 ) 2 ) ∣ a b = ( 3 4 7 ) 2 3 5 1 1 0 3 8 7 2 ( 2 π ) =
( 3 4 7 ) 2 3 5 5 5 1 9 3 6 π = β 2 3 λ α π ⟹ α + β + λ = 5 5 2 2 8 8 .