Given equation of asymptotes
x+y=−3‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)and
‌‌2x−y=−1‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)On solving Eqs. (i) and (ii), we get centre of hyperbola i.e.
x=‌,y=‌i.e. Centre
≡(‌,‌)Also combined equation of asymptotes is
(x+y+3)(2x−y+1)=0⇒‌‌2x2−y2+xy+7x−2y+3=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)Now, equation of hyperbola is (when equation of asymptotes are known) is
(x+y+3)(2x−y+1)+c=0where
cis constant.
∵ (iii) passes through
(1,−2).
So,
(1−2+3)(2+2+1)+c=0⇒‌‌(2×5)+c=0⇒c=−10∴ Equation of hyperbola is
(x+y+3)(2x−y+1)−10=0Also, equation of conjugate hyperbola is (as we know that, equation of hyperbola, conjugate hyperbola, and
pair of asymptotes are same but only difference in constant term).
=2x2+xy−y2+7x−2y+13=0