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Usually in simple cases setting the constant term in the equation of a hyperbola yields the ssymptotes e.g., the hyperbolas $x^2-y^2/b^2=1$ and $(2x+y-+1)(x-3y+2)=1$ have asymptotes as $x^2/a^2-y^2/b^2=0$ and $2x+y+1=0, x-3y+2=0$, respectively.

In general for a hyperbola $$ax^2+by^2+2hxy+2gx+2fy+c=0~~~~(1)$$ on chnges $c$ to $c'$, so that the given quadratic represents pair of straight lines as per the condition that $$abc'+2fgh-af^2-bg^2-c'h^2=0~~~(2)$$ Next. by replacing $c$ by $c'$ in (1) one is supposed to separate the pair of straight lines which finally give the asymptotes.

one can see some more ways in Finding the asymptotes of a general hyperbola

I wonder if there could be a more direct and simple method for doing this. One may help me by illustrating the better method for this case: $$3x^2-y^2+2xy+7x-y-1=0.$$

Dharmendra Singh
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1 Answers1

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You may put $y=mx+c$ in the equation of thr curve and separate out the terms of different powers of $x$ and set their coefficients to zero leaving out the constant term. For your example $$3x^2-y^2+2xy+7x-y-1=0.~~~~~(1)$$ you get $$(3+2m-m^2)x^2+(7-m+2c-2mc)x+(-1-c^2-c)=0~~~~~~(2)$$ Now set $$(3+2m=m^2)=0, ~~ (7-m+2c-2mc)=0.$$

First Eq. gives $m=3,-1$ the second equation gives $c=1$ for $m=3$ and $c=-2$ for $m=-1$. $y=3x+1$ and $y=-x-2$.

Note that the obtained $c$ values are for the asymptotes, these will not satisfy $(-1-c^2-c)=0$ because Eq. (2) corresponds to the hyperbola and not to the asymptotes.

Z Ahmed
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