Question

For a hyperbola whose centre is at (1, 2) if asymptotes are parallel to lines 2x + 3y = 0 and x + 2y = 1, then equation of hyperbola passing through (2, 4) is

Answer 1

as the asymptotes are parallel to the lines

$2x + 3y = 0 \: \: \: \: \: and \\ x + 2y = 1$

so, there equations are

$2x + 3y + \alpha = 0 \: \: \: \: \: and \\ x + 2y + \beta = 0 \\ also \: these \: asymptotes \: passes \: through \: the \: centre \: of \: hyperbola \: \\ so \: values \: of \: \alpha \: and \: \beta can \: be \: found \\ \\ therefore \\ \alpha = - 8 \: \: \: and \\ \beta = - 5 \\ so \: eq. \: of \: asymptotes \: comes \: out \: to \: be \\ 2x + 3y - 8 = 0 \: \: \: and \\ x + 2y - 5 = 0 \\ and \: their \: joint \: equation \: is \: \\ (2x + 3y - 8)(x + 2y - 5) = 0 \\ \\ since \: eq. \: of \: asymptotes \: and \: hyperbola \: differ \: only \\ in \: constant \: term \\ so \: eq. \: of \: hyperbola \: is \: (2x + 3y - 8)(x + 2y - 5) + c = 0 \\ and \: it \: passes \: through \: (2 \: \: . \: \: 4) \\ on \: putting \: it \: we \: get \\ ((2.2) + (3.4) - 8)((2) + (2.4) - 5) + c = 0 \\ (4 + 12 - 8)(2 + 8 - 5) + c = 0 \\ (8 \times 5) + c = 0 \\ c = - 40 \\ \\ \\ so \: required \: eq. \: is \: \\ (2x + 3y - 8)(x + 2y - 5) - 40 = 0$
hope it helps（＾＿－）≡★