0
$\begingroup$

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 .

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.