# [Math] the eccentricity of the hyperbola given by $3x^2+7xy+2y^2-11x-7y+10=0$

analytic geometryconic sectionsgeometry

What is the eccentricity of the hyperbola given by the following equation?
$$3x^2+7xy+2y^2-11x-7y+10=0$$

what i try

$$S = 3x^2+7xy+2y^2-11x-7y+10$$

$$\displaystyle \frac{dS}{dx}=6x+7y-11=0$$ and $$\displaystyle \frac{dS}{dy}=7x+4y-7=0$$

center of hyperbola is at $$\displaystyle x = 1/5$$ and $$y=7/5$$

put $$x=X-1/5$$ and $$y=y-7/5$$ into $$S$$

$$\displaystyle 3\bigg(X-1/5\bigg)^2+7\bigg(X-1/5\bigg)\bigg(Y-7/5\bigg)+2\bigg(Y-7/5\bigg)^2-11\bigg(X-1/5\bigg)-7\bigg(Y-7/5\bigg)+10=0$$

$$3X^2+2Y^2+7XY+611=0$$

How do i solve it help me please

Rewrite the equation of that hyperbola as follows: $$(x+2y-3)(3x+y-2)=-4.$$ Lines $$x+2y-3=0$$ and $$3x+y-2=0$$ are then the asymptotes, forming between them an angle $$\theta$$ given by $$\cos\theta={(1,2)\cdot(3,1)\over|(1,2)|\cdot|(3,1)|}={1\over\sqrt2}.$$ If $$a$$ and $$b$$ are the hyperbola semi-axes, the eccentricity $$e$$ is then given by $$e=\sqrt{1+b^2/a^2}=\sqrt{1+\tan^2(\theta/2)}={1\over\cos(\theta/2)}=\sqrt{2\over1+\cos\theta}=2\sqrt{1-1/\sqrt2}.$$