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The question goes like:

A general hyperbola has its axes at right angle and the asymptotes have reflection symmetry about these axes. However, the axes may be rotated, unlike the standard form of a hyperbola. The general equation of a hyperbola can be written as $ax^2+bxy+cy^2+dx+ey+f=0.$

The general form of a hyperbola passing through the point $(x_0,y_0)$ can be derived from the asymptotes using the hyperbola equations. $(A_1x+B_1y+C_1)(A_2x+B_2y+C_2)=(A_1x_0+B_1y_0+C_1)(A_2x_0+B_2y_0+C_2)$ where $A_1x+B_1y+C_1=0$ and $A_2x+B_2y+C_2=0$ are the two asymptotes.

Find the general equation of a hyperbola which passes through the point $(1,1)$ has an asymptote $y=0$ and has an axis of symmetry $y=2x+2$.

Any help will be really appreciated- thank you!

Hosam Hajeer
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Yeonsu Na
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1 Answers1

3

From the reflective property of the asymptotes with respect to the axes of the hyperbola, we can find the equation of the other asymptote.

The two asymptotes and the axes of symmetry all meet in a single point. And this point is the center of symmetry of the hyperbola.

We're given that $y=0$ is an asymptote, and that $y = 2 x + 2 $ is an axis. These two lines intersect at $ x = -1, y = 0 $, so the center of the hyperbola is the point $(-1, 0)$.

Now we want to refect the line $y = 0$ about the line $ y = 2 x + 2 $, a small sketch will reveal that the reflected line will have a slope of $ \tan( 2 \tan^{-1}(2) ) = \dfrac{2(2)} {(1 - 4} = - \dfrac{4}{3} $

Hence the reflected line (the other asymptote) equation is $ y = -\dfrac{4}{3} x + b $

Since the hyperbola center $(-1,0)$ is on this reflected line, then $ b = -\dfrac{4}{3} $. Hence, the equation of the other asymptote is $ y = - \dfrac{4}{3} (x + 1) $

From here, we just apply the formula given in the question to find the equation of the hyperbola, using the point $(x_0, y_0) = (1, 1) $

$ ( y ) (\dfrac{4}{3} x + y + \dfrac{4}{3} ) = (1) (\dfrac{4}{3}(1) + (1) + \dfrac{4}{3} ) $

And this simplifies to

$ ( y ) (\dfrac{4}{3} x + y + \dfrac{4}{3} ) = \dfrac{11}{3}$

Multiplying the brackets on the left side and multiplying through by $(3)$, results in the final form of the equation of the hyperbola

$ 4 xy + 3 y^2 + 4 y - 11 = 0 $

Hosam Hajeer
  • 21,978
  • Thanks so much, I haven't done too much conics so I am a bit bad. So this really helps. Thanks again. – Yeonsu Na Sep 22 '22 at 19:53
  • You're welcome. My pleasure. – Hosam Hajeer Sep 22 '22 at 20:00
  • I think part of the difficulty was that the original wording of the question, "the asymptotes have reflection symmetry about these axes", is ambiguous. It seems to say that each asymptote is symmetric about the axes, but that's impossible. Instead, each asymptote is the reflection of the other, as explained in this answer. If you look at both asymptotes simultaneously, they form a big X that is symmetric across either axis of the hyperbola. – David K Sep 22 '22 at 21:07
  • Just to clarify, the axes they are referring to are the transverse and conjugate axes right? – Yeonsu Na Sep 22 '22 at 22:22
  • Yes. You're right. – Hosam Hajeer Sep 22 '22 at 22:47