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

## Best Answer

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 $$ e=\sqrt{1+b^2/a^2}=\sqrt{1+\tan^2(\theta/2)}={1\over\cos(\theta/2)}=\sqrt{2\over1+\cos\theta}=2\sqrt{1-1/\sqrt2}. $$