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How to find the eccentricity of this conic?

$$4(2y-x-3)^2-9(2x+y-1)^2=80$$

My approach :

I rearranged the terms and by comparing it with general equation of 2nd degree, I found that its a hyperbola. Since this hyperbola is not in standard form $x^2/a^2-y^2/b^2= 1$, I don't know how to find its eccentricity.

Please guide me.

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  • $\begingroup$ The eccentricity should be preserved by rigid motions, right? $\endgroup$
    – anon
    Jun 30, 2012 at 7:55
  • $\begingroup$ What level are you asking this question from,... is this a pre-calculus course? Just so the responses and terminology would be more appropriate for your level. If so, you might want to tag your question with pre-calculus or the related as well. $\endgroup$
    – night owl
    Jun 30, 2012 at 8:29

3 Answers 3

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First make the following change of coordinates: $$ u=\frac{x-2y}{\sqrt{3}}, \ v=\frac{2x+y}{\sqrt{3}}. $$ With these coordinates the canonical basis $e_1=(1,0), e_2=(0,1)$ is transformed into $e_1'=\frac{(1,2)}{\sqrt{3}}, e_2'=\frac{(-2,1)}{\sqrt{3}}$ which is clearly an orthonormal basis. The equation now reads: $$ 4(-\sqrt{3}u-3)^2-9(\sqrt{3}v-1)^2=80, $$ i.e. $$ \frac{(u+\sqrt{3})^2}{a^2}-\frac{(v-1/\sqrt{3})^2}{b^2}=1. $$ with $$ a^2=20/3>b^2=80/27. $$ So, the eccentricity is $$ e=\sqrt{1+b^2/a^2}=\sqrt{1+4/9}=\sqrt{13}/3. $$

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HINT: The presence of $x y $ term shows that the axes are not going to be nicely parallel to x- and y- axes. There is a formula like $ \tan 2 \theta = 2B/(A-C) $ to find angle of rotation to align along x- and y-.

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Note that for a standard hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the individual terms of this equation can be represented as perpendicular distances of the point from the x-axis ($y=0$) and the y-axis ($x=0$). This is because, for a point, the x-coordinate is the perpendicular distance from the y-axis and the y-coordinate is the perpendicular distance from the x-axis.

Alternate equation of hyperbola: $$\frac{\text{(Distance from conjugate axis)}^2}{a^2} - \frac{\text{(Distance from transverse axis)}^2}{b^2} = 1$$ (as $x=0$ and $y=0$ are the conjugate and transverse axes respectively)

In this problem, we have $4(2x-y-3)^2 - 9(2x+y+1)^2=1$ as the hyperbola. Let $2y-x-3$ and $2x+y-1$ be the conjugate and transverse axes respectively. The distances of a point $(x,y)$ from these lines are $d_c = \frac{|2y-x-3|}{\sqrt{5}}$ and $d_t = \frac{|2x+y-1|}{\sqrt{5}}$ respectively.

Now rewriting the original equation in terms of $d_c$ and $d_t$: $$4{d_c}^2-9{d_t}^2=16$$ which is $$\frac{{d_c}^2}{4}-\frac{{d_t}^2}{\frac{16}{9}}=1$$

which gives us the effective values of $a=2$ and $b=4/3$. We can now calculate the eccentricity.

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    $\begingroup$ The necessary precondition for this to work is that the given lines have to perpendicular, as only then can we try to assume them as axes. $\endgroup$
    – zxen
    Dec 20, 2023 at 16:36

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