The equations of the asymptotes of a hyperbol | vedclass

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(B) The equation of the hyperbola is given by $(x+y+3)(2x-y+1) + \lambda = 0$. Since the point $(1,-2)$ lies on the hyperbola,we substitute $x=1$ and

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$2x^2+xy-y^2+7x-2y+13=0$

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(B) The equation of the hyperbola is given by $(x+y+3)(2x-y+1) + \lambda = 0$. Since the point $(1,-2)$ lies on the hyperbola,we substitute $x=1$ and $y=-2$: $(1-2+3)(2(1)-(-2)+1) + \lambda = 0$ $(2)(5) + \lambda = 0 \implies \lambda = -10$. Thus,the equation of the hyperbola is $(x+y+3)(2x-y+1) - 10 = 0$. Expanding this: $2x^2 - xy + x + 2xy - y^2 + y + 6x - 3y + 3 - 10 = 0$,which simplifies to $2x^2 + xy - y^2 + 7x - 2y - 7 = 0$. The equation of the conjugate hyperbola is $(x+y+3)(2x-y+1) + \lambda' = 0$. Since the conjugate hyperbola passes through the point symmetric to $(1,-2)$ with respect to the center of the hyperbola,or by using the property that the constant term changes such that the sum of the constants of the hyperbola and its conjugate is $2 	imes (	ext{constant of the product of asymptotes})$,we find the constant term. Alternatively,the equation of the conjugate hyperbola is $(x+y+3)(2x-y+1) + 10 = 0$. Expanding this: $2x^2 + xy - y^2 + 7x - 2y + 3 + 10 = 0$,which simplifies to $2x^2 + xy - y^2 + 7x - 2y + 13 = 0$.

The equations of the asymptotes of a hyperbola are $x+y+3=0$ and $2x-y+1=0$. If $(1,-2)$ is a point on this hyperbola,find the equation of its conjugate hyperbola.

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