Hyperbola

Given equation of asymptotes
x+y=−3‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
and‌‌2x−y=−1‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
On solving Eqs. (i) and (ii), we get centre of hyperbola i.e.x=‌

−4

3

,y=‌

−5

3

i.e. Centre≡(‌

−4

3

,‌

−5

3

)
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=0
wherecis 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=0
Also, 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