1 7 x 2 + 2 2 y 2 + ( ∫ 0 ∞ 2 t 4 + t 2 + 1 1 d t ) x + ( ∫ 0 1 2 z ( 1 − z ) e 2 z ( 1 − z ) d z ) y + ( 2 5 ! ) ! = 0
Find the eccentricity of the ellipse satisfying the equation above. Give your answer up to three decimal places.
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Yes the short trick works. The integrals are indeed just a distraction. Well done!
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Short trick
e = 1 − c o e f . o f y 2 c o e f . o f x 2
Where c o e f . o f x 2 < c o e f . o f y 2
Proof
The general equation of the above ellipse will definitely be of the form
1 7 ( x − x 1 ) 2 + 2 2 ( y − y 1 ) 2 = c
For some constants x 1 , x 2 , c
Rewriting in our general form
a 2 x 2 + b 2 y 2 = 1
1 7 c ( x − x 1 ) 2 + 2 2 c ( y − y 1 ) 2 = 1
And e = 1 − b 2 a 2
When you'll substitute values in formula for "e" you'll see that the constant 'c' cancels out, this explains why the short trick is valid.
And no need to worry about the integrals, it is clear that they will converge.