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The second asymptote will be of the form $$2x-y+\lambda=0$$

because it is a rectangular hyperbola. I know it will pass through the centre of the hyperbola. But how do I find it?

One guess was that we could use partial derivatives, but that method is used for pair of straight lines, while this is a hyperbola. I could really use some insight.

Aditya
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2 Answers2

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Comparing the given quadratic with $$Ax^2+2Hxy+By^2+2Gx+2Fy+C=0~~~(1)$$ We get $$A=2, H=3/2, B=-2, G=-3, F=13/2, C=-36~~~~(2)$$ The combined eq. of the asymptotes of the given hyperbola would be $$2x^2+3xy-2y^2-6x+13y+D=0~~~(3)$$, provided (3) represents a pair of straight line. The condition for this is $$ABD+2FGH-AF^2-BG^2-DH^2=0~~~~(3)$$ $$\implies -4D-117/2-169/2+18-9D/4=0 \implies D=-20.~~~~(4)$$ So the combined Eq. of asymptotes is $$2x^2+3xy-2y^2-6x+13y-20=0 \implies~(x+2y-5)(2x-y+4)=0~~(5)$$ Finally, we get $\lambda=4.$

José Carlos Santos
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Z Ahmed
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You can, in fact, use partial derivatives to find the center of any conic that has a center. See this answer for a general explanation of why that works. In the particular case of a hyperbola, as already noted elsewhere the equations of the family of hyperbolas that share asymptotes can be written in the form $(asymptote_1)(asymptote_2)=const$ for various constants. A pair of intersecting lines is the degenerate member of this family for which the constant is zero. When you differentiate this, the constant term drops out, so you get the same center point for all of the hyperbolas in the family, as one might expect. So, you can proceed as you already know how to do for a pair of intersecting lines: set the partial derivatives to zero and solve the system for $x$ and $y$, then use that value to determine $\lambda$.

You can avoid all of that by using the above equation schema directly: divide $2x^2+3xy-2y^2-6x+13y-36$ by $x+2y-5$. The quotient will be a linear term that represents the other asymptote in the product on the left-hand side of the schema, while the remainder will be the negative of the constant on the right-hand side. Since you already know that the other asymptote has an equation of the form $2x-y+\lambda=0$, you can expand $(x+2y-5)(2x-y+\lambda)$ and compare coefficients to the hyperbola’s equation to find $\lambda$ and the left-over constant term. In this case, the problem’s author was nice to you and didn’t thrown in a scalar multiple of one of these equations when forming the equation of the hyperbola, but in general you’ll need to account for that if you “reverse-engineer” the equation from the asymptotes.

amd
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  • So we can use partial derivatives to find the centre of the hyperbola and simply pass the given asymptote through it, avoiding all the multiplication of the equations. – Aditya Mar 30 '20 at 03:55
  • @Aditya Just so, but only because in this case we know that the asymptotes are perpendicular so that we can get most of the second asymptote’s equation by inspection. Factoring the conic equation works in all cases. Note, too, that you can factor only the quadratic part of the conic equation to get the directions of the asymptotes and then pass those through the center, which is sometimes easier to do. – amd Mar 30 '20 at 17:22
  • Well, 'reverse engineering' a hyperbola is incredibly tough – Aditya Mar 30 '20 at 17:32
  • @Aditya I’d say tedious rather than tough, myself :) You can find the directions of the asymptotes by computing its intersections with the line at infinity, which can be done purely mechanically. Finding the center is also a mechanical process. From there it’s just a matter of working out what the left-over constant term is, which again is a largely mechanical process. – amd Mar 30 '20 at 18:25