Let x+y+1=0 and x-y+4=0 be the asymptotes of | vedclass

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(A) The asymptotes of a hyperbola are given by the equations $x+y+1=0$ and $x-y+4=0$. Solving these equations simultaneously gives the center of the h

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$4 \sqrt{3}$

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(A) The asymptotes of a hyperbola are given by the equations $x+y+1=0$ and $x-y+4=0$. Solving these equations simultaneously gives the center of the hyperbola: $x+y=-1$ and $x-y=-4$. Adding them gives $2x = -5$,so $x = -2.5$. Subtracting gives $2y = 3$,so $y = 1.5$. Thus,the center is $(-2.5, 1.5)$. The equation of the hyperbola is $(x+2.5+y-1.5)(x+2.5-(y-1.5)) = \lambda$,which simplifies to $(x+y+1)(x-y+4) = k$. Since the point $(1,1)$ lies on the hyperbola,we substitute $x=1$ and $y=1$: $(1+1+1)(1-1+4) = k \Rightarrow 3 	imes 4 = 12$. So $k=12$. The equation is $(x+2.5)^2 - (y-1.5)^2 = 12$. Comparing this to $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$,we have $a^2 = 12$ and $b^2 = 12$,so $a = \sqrt{12} = 2\sqrt{3}$ and $b = 2\sqrt{3}$. The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 	imes 12}{2\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$.

Let $x+y+1=0$ and $x-y+4=0$ be the asymptotes of a hyperbola $H$. If $(1,1)$ is a point on $H$,then the length of the latus rectum of $H$ is

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