A hyperbola whose asymptotes are $x+2y+3=0$ and $3x+4y+5=0$ , is passing through the point (1,-1) . Then we have to find the equation of the conjugate hyperbola .

I thought about it alot .. but could not get any start .

can anybody provide me a hint .

  • $\begingroup$ Once you insert your point to get $k$, David Quinn's hint gives the answer as a general quadratic form $Ax^2 + Bxy + Cy^2 + Dx + E y + F = 0$. From that form, you get both the hyperbola and the conjugate hyperbola which have symmetry axes which are tilted by $\phi$. If you need the angles $\phi$, these can be computed from $\cot 2 \phi = (A-C)/B$. Once you rotate your coordinate system by $-\phi$, you get standard hyperbola equations $(x' - x_0')^2/a^2 - (y' - y_0')^2/b^2 = 1$. $\endgroup$ – Andreas Apr 28 '17 at 15:02

hint...The hyperbola you want is $(x+2y+3)(3x+4y+5)=k$ For a suitable choice of $k$

  • $\begingroup$ How you got this ? $\endgroup$ – Koolman Apr 28 '17 at 16:40
  • $\begingroup$ Consider what are the asymptotes of the family of hyperbolae $(x+y)(x-y)=k$ and why? $\endgroup$ – David Quinn Apr 28 '17 at 16:46
  • $\begingroup$ Is this applicable for every hyperbola $\endgroup$ – Koolman Apr 28 '17 at 16:51

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