Identifying an ellipse

Geometry Level pending

You are given four points: A ( 0 , 6 ) , B ( 3 , 9 ) , C ( 6 , 8 ) , D ( 8 , 3 ) A(0, 6), B(3, 9), C(6, 8), D(8, 3) , and you are asked to construct an ellipse that passes through these four points and at the same time be tangent to the x x -axis. It turns out there are two such ellipses. Find the sum S S of their areas, and enter 1 0 4 S \lfloor 10^4 S \rfloor as your answer.


The answer is 1586688.

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1 solution

Yuriy Kazakov
Apr 9, 2021

I don't find analytic solution for problem. I use Wolframalpha.

Find ellipse equation in form

a x 2 + b y 2 + c x y + d x + f y = 1 a \cdot x^2+b \cdot y^2+c \cdot x \cdot y+d \cdot x+f \cdot y=1

For
y = 0 y=0

a x 2 + d x = 1 a \cdot x^2+d \cdot x=1

D = d 2 + 4 a = 0 D=d^2+4a=0

a = d 2 4 a=-\frac{d^2}{4}

d 2 4 x 2 + b y 2 + c x y + d x + f y = 1 -\frac{d^2}{4} \cdot x^2+by^2+c\cdot x \cdot y+d \cdot x+f \cdot y=1

Four equations for points A ( 0 , 6 ) , B ( 3 , 9 ) , C ( 6 , 8 ) , D ( 8 , 3 ) A(0,6),B(3,9),C(6,8),D(8,3)

b 6 2 + f 6 = 1 b \cdot 6^2+f \cdot 6=1

d 2 4 3 2 + b 9 2 + c 3 9 + d 3 + f 9 = 1 -\frac{d^2}{4} \cdot 3^2+b \cdot 9^2+c \cdot 3 \cdot 9+d \cdot 3+f \cdot 9=1

d 2 4 6 2 + b 8 2 + c 6 8 + d 6 + f 8 = 1 -\frac{d^2}{4} \cdot 6^2+b \cdot 8^2+c \cdot 6 \cdot 8+d \cdot 6+f \cdot 8=1

d 2 4 8 2 + b 3 2 + c 8 3 + d 8 + f 3 = 1 -\frac{d^2}{4} \cdot 8^2+b \cdot 3^2+c \cdot 8 \cdot 3+d \cdot 8+f \cdot 3=1

Solve system

Properties of ellipse 1

Properties of ellipse 2

ellipse1

area enclosed | 102.0215377764767286

ellipse2

area enclosed | 56.64732607727403969317

102.0215377764767286+56.64732607727403969317

Use Wolframalpha

1586688

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