The equation of the hyperbola whose asymptote | vedclass

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(B) The equation of the pair of asymptotes is given by the product of the lines: $(3x+4y-2)(2x+y+1)=0$. Since the equation of the hyperbola differs fr

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$6x^2+11xy+4y^2-x+2y-22=0$

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(B) The equation of the pair of asymptotes is given by the product of the lines: $(3x+4y-2)(2x+y+1)=0$. Since the equation of the hyperbola differs from the equation of its asymptotes only by a constant $\lambda$,we can write: $(3x+4y-2)(2x+y+1)=\lambda$. Given that the hyperbola passes through the point $(1,1)$,we substitute $x=1$ and $y=1$ into the equation: $(3(1)+4(1)-2)(2(1)+1+1)=\lambda$. $(5)(4)=\lambda$,which gives $\lambda=20$. Substituting $\lambda=20$ back into the equation: $(3x+4y-2)(2x+y+1)=20$. Expanding the left side: $6x^2+3xy+3x+8xy+4y^2+4y-4x-2y-2=20$. Simplifying: $6x^2+11xy+4y^2-x+2y-2=20$. Thus,the equation is $6x^2+11xy+4y^2-x+2y-22=0$.

The equation of the hyperbola whose asymptotes are the lines $3x+4y-2=0$ and $2x+y+1=0$ and which passes through the point $(1,1)$ is

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