It's all Ellipses!

Geometry Level 5

If the area A A of the ellipse above can be represented as A = α π β 3 2 λ A = \dfrac{\alpha\pi}{\beta^{\frac{3}{2}}\sqrt{\lambda}} , where α , β \alpha, \beta and λ \lambda are coprime positive integers, find α + β + λ \alpha + \beta + \lambda .


The answer is 552288.

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1 solution

Rocco Dalto
Sep 15, 2018

Using a x 2 + b x y + c y 2 + d x + e y = 0 ax^2 + bxy + cy^2 + dx + ey = 0 and the points I chose above to generate the ellipse.

(1) ( 0 , 4 ) : 4 c e = 0 c = e 4 (0,-4): 4c - e = 0 \implies \boxed{c = \dfrac{e}{4}}

(2) ( 8 , 4 ) : 16 a + 8 b + 4 c + 2 d + e = 0 (8,4): 16a + 8b + 4c + 2d + e = 0

(3) ( 8 , 4 ) : 16 a 8 b + 4 c + 2 d e = 0 (8,-4): 16a - 8b + 4c + 2d - e = 0

Subtracting (3) from (2) we obtain: 8 b + e = 0 b = e 8 8b + e = 0 \implies \boxed{b = -\dfrac{e}{8}}

(4) ( 5 , 8 ) : 25 a + 40 b + 64 c + 5 d + 8 c = 0 (5,8): 25a + 40b + 64c + 5d + 8c = 0

Replacing c = e 4 c = \dfrac{e}{4} and b = e 8 b = -\dfrac{e}{8} into (2) and (4) \implies

16 a + 2 d = e 16a + 2d = -e

25 a + 5 d = 19 e 25a + 5d = -19e

a = 11 e 10 \implies \boxed{a = \dfrac{11e}{10}} and d = 93 e 10 \boxed{d = -\dfrac{93e}{10}}

11 10 x 2 1 8 x y + 1 4 y 2 93 10 x + y = 0 \implies \dfrac{11}{10}x^2 - \dfrac{1}{8}xy + \dfrac{1}{4}y^2 - \dfrac{93}{10}x + y = 0 \implies 44 x 2 5 x y + 10 y 2 372 x + 40 y = 0 10 y 2 + 5 ( x 8 ) y + 44 x 2 372 x = 0 44x^2 - 5xy + 10y^2 - 372x + 40y = 0 \implies 10y^2 + 5(x - 8)y + 44x^2 - 372x = 0

Solving for y y we obtain:

y = x 8 4 ± 1 20 5 347 2207744 ( 347 x 1448 ) 2 y = \dfrac{x - 8}{4} \pm \dfrac{1}{20}\sqrt{\dfrac{5}{347}}\sqrt{2207744 - (347x - 1448)^{2}} .

y ( x ) = x 8 4 + 1 20 5 347 2207744 ( 347 x 1448 ) 2 y(x) = \dfrac{x - 8}{4} +\dfrac{1}{20}\sqrt{\dfrac{5}{347}}\sqrt{2207744 - (347x - 1448)^2} for the portion of the ellipse above the line y = x 8 4 y = \dfrac{x - 8}{4} .

Setting 2207744 ( 347 x 1448 ) 2 = 0 x = 1448 ± 448 11 347 2207744 - (347x - 1448)^2 = 0 \implies x = \dfrac{1448 \pm 448\sqrt{11}}{347} are the points of intersection of the ellipse and the line x 8 4 \dfrac{x - 8}{4} .

Letting a = 1448 448 11 347 a = \dfrac{1448 - 448\sqrt{11}}{347} and b = 1448 + 448 11 347 b =\dfrac{1448 + 448\sqrt{11}}{347} the area of the ellipse is A = 2 a b y ( x ) x 8 4 d x = A = 2\displaystyle\int_{a}^{b} y(x) - \dfrac{x - 8}{4} \:\ dx =
1 10 5 347 a b 2207744 ( 347 x 1448 ) 2 d x = \dfrac{1}{10}\sqrt{\dfrac{5}{347}}\displaystyle\int_{a}^{b} \sqrt{2207744 - (347x - 1448)^{2}} dx = 1 2 5 347 a b 2207744 ( 347 x 1448 ) 2 d x \dfrac{1}{2\sqrt{5}\sqrt{347}}\displaystyle\int_{a}^{b} \sqrt{2207744 - (347x - 1448)^2} dx .

For I ( x ) = 2207744 ( 347 x 1448 ) 2 d x I(x) = \displaystyle\int \sqrt{2207744 - (347x - 1448)^2} dx

Let 347 x 1448 = 2207744 sin ( θ ) d x = 2207744 347 cos ( θ ) 347x - 1448 = \sqrt{2207744}\sin(\theta) \implies dx = \dfrac{\sqrt{2207744}}{347}\cos(\theta)

I ( θ ) = 2207744 347 cos 2 ( θ ) d θ = 2207744 694 ( 1 + cos ( 2 θ ) ) d θ = 2207744 694 ( θ + sin ( θ ) cos ( θ ) ) \implies I(\theta) = \dfrac{2207744}{347}\displaystyle\int \cos^2(\theta) d\theta = \dfrac{2207744}{694}\displaystyle\int (1 + \cos(2\theta)) d\theta = \dfrac{2207744}{694}(\theta +\sin(\theta)\cos(\theta))

I ( x ) = 2207744 694 ( arcsin ( 347 x 1448 2207744 ) + 347 x 1448 2207744 2207744 ( 347 x 1448 ) 2 ) \implies I(x) = \dfrac{2207744}{694}(\arcsin(\dfrac{347x - 1448}{\sqrt{2207744}}) + \dfrac{347x - 1448}{2207744}\sqrt{2207744 - (347x - 1448)^2}) \implies

A = 551936 ( 347 ) 3 2 5 ( arcsin ( 347 x 1448 2207744 ) + 347 x 1448 2207744 2207744 ( 347 x 1448 ) 2 ) a b = 1103872 ( 347 ) 3 2 5 ( π 2 ) = A = \dfrac{551936}{(347)^{\frac{3}{2}}\sqrt{5}}(\arcsin(\dfrac{347x - 1448}{\sqrt{2207744}}) + \dfrac{347x - 1448}{2207744}\sqrt{2207744 - (347x - 1448)^2})|_{a}^{b} = \dfrac{1103872}{(347)^{\frac{3}{2}}\sqrt{5}}(\dfrac{\pi}{2}) =

551936 π ( 347 ) 3 2 5 = α π β 3 2 λ \dfrac{551936\pi}{(347)^{\frac{3}{2}}\sqrt{5}} = \dfrac{\alpha\pi}{\beta^{\frac{3}{2}}\sqrt{\lambda}} α + β + λ = 552288 \implies \alpha + \beta + \lambda = \boxed{552288} .

omg and i thought that i would be able to solve this. If u came up with this , then.....hats off man

Anuj Tripathi - 2 years, 8 months ago

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I created the problem. Thanks.

Rocco Dalto - 2 years, 8 months ago

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How can I follow or subscribe to u. Btw are u an expert in conic section or do u just love it. I am really not that good in conic section. Still thank u for ur active cooperation in the community and for ur problems. :)

Anuj Tripathi - 2 years, 8 months ago

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