Question outta nowhere

Geometry Level 5

17 x 2 + 22 y 2 + ( ∫ 0 ∞ 1 2 t 4 + t 2 + 1 d t ) x + ( ∫ 0 1 2 z ( 1 − z ) e 2 z ( 1 − z ) d z ) y + ( 25 ! ) ! = 0 17x^{2}+22y^{2}+\displaystyle \left(\int^{\infty}_{0} \frac{1}{2t^{4}+t^{2}+1} \ dt \right)x \\ +\displaystyle \left(\int^{1}_{0} 2z(1-z)e^{2z(1-z)} \ dz \right)y+ \left(25!\right)!=0

Find the eccentricity of the ellipse satisfying the equation above. Give your answer up to three decimal places.


The answer is 0.477.

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Krishna Sharma
Apr 20, 2015

Short trick

e = 1 − c o e f . o f x 2 c o e f . o f y 2 \displaystyle e = \sqrt{ 1 - \dfrac{ coef. of x^2}{ coef. of y^2}}

Where c o e f . o f x 2 < c o e f . o f y 2 coef. of x^2 < coef. of y^2

Proof

The general equation of the above ellipse will definitely be of the form

17 ( x − x 1 ) 2 + 22 ( y − y 1 ) 2 = c \displaystyle 17(x - x_1)^2 + 22(y-y_1)^2 = c

For some constants x 1 , x 2 , c x_1, x_2, c

Rewriting in our general form

x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1

( x − x 1 ) 2 c 17 + ( y − y 1 ) 2 c 22 = 1 \displaystyle \dfrac{(x-x_1)^2}{\frac{c}{17}} + \dfrac{(y-y_1)^2}{\frac{c}{22}} = 1

And e = 1 − a 2 b 2 e = \sqrt{1 - \dfrac{a^2}{b^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.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Yes the short trick works. The integrals are indeed just a distraction. Well done!

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