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I need to Find the area of conic section given by:

$5x^2 - 6xy + 5y^2 = 8$, Using substitution $x = u+ v$ and $y = u-v$ we get :

$5\left(u+v\right)^2 - 6\left(u^2 - v^2\right) + 5\left(u-v\right)^2 = 8$

which simplifies to :

$4u^2 + 16v^2 = 8$

thus, this is an equation of ellipse, $\dfrac{u^2}{2} + \dfrac{v^2}{1/2} = 1$,

Area is given by : $\pi a b = \pi \sqrt{2} \sqrt{1/2} = \pi$

Is My Solution to given problem correct ?

Thank you

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    Does your substitution preserve lengths? – CY Aries Sep 14 '19 at 17:20
  • @CY Aries: I'm sorry I cannot understand your hint, can you please elaborate ? –  Sep 14 '19 at 17:26
  • The distance between the points $(x,y)=(0,0)$ and $(x,y)=(1,0)$ is $1$. The two points are corresponding to $(u,v)=(0,0)$ and $(u,v)=(0.5, 0.5)$. The distance between $(0,0)$ and $(0.5, 0.5)$ is $1/\sqrt{2}$. The lengths change after applying the substitution. So do areas. – CY Aries Sep 14 '19 at 17:30
  • So , I need to multiply my area by the "scaling" Factor ie the value of $\left|J\right|$,where J is the Jacobian of the Transformation $x$ and $y$, Since the value of Jacobian is $2$ (After neglecting the signs), the area should be doubled ? –  Sep 14 '19 at 17:38
  • Is my reasoning correct ? –  Sep 14 '19 at 17:38
  • See the first part of this answer for a way to compute this area directly from the original equation. – amd Sep 14 '19 at 17:58

1 Answers1

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[Adding this answer in order that yet another question that was answered in a comment not remain formally unanswered.]

As CY Aries points out in a comment, your transformation isn’t area-preserving. You must either rescale the area computed from the transformed equation by multiplying by the Jacobian determinant of the transformation (as you noted), or use an area-preserving transformation in the first place. The one you used originally is a similarity, so it’s easily adjusted to preserve areas: normalize it to $x={u+v\over\sqrt2}$, $y={u-v\over\sqrt2}$. Either way you’ll end up with twice the area that you computed with your original transformation.

amd
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