Hallo Bertie,
60y2+65x2−128xy−448x+448y+768+60⋅c=0c>0 hyperbola between lines
Here is the answer:
I.
1. line13x−10y−48=0y=m1x+b1y=1310x−4810m1=1310b1=−4810or m1x+b1−y=0(1310x−4810−y=0)2. line5x−6y−16=0y=m2x+b2y=56x−166m2=56b2=−166or m2x+b2−y=0(56x−166−y=0)Intersection 1. lineand 2.linexc=b2−b1m1−m2=327yc=m1b2−b1m2m1−m2=87
II. The Hyperbola is the product of the two lines + c : (m1x+b1−y)(m2x+b2−y)+c=0 c > 0
or y2+m1m2x2+(m1b2+m2b1)x−(m1+m2)xy−(b1+b2)y+b1b2+c=0 or y2+1312x2−11215x−3215xy+11215y+645+c=0⋅60or 60y2+65x2−448x−128xy+448y+768+60⋅c=0
III. The Hyperbola is: Y2+aX2+bXY+c=0.|X=x−xc, Y=y−yc[see #6]
a=m1m2=1313[see #11]b=−(m1+m2)=−3215[see #11]xc=b2−b1m1−m2=327yc=m1b2−b1m2m1−m2=87
let us put all things together:
(y−m1b2−b1m2m1−m2)2+m1⋅m2⋅(x−b2−b1m1−m2)2−(m1+m2)⋅(x−b2−b1m1−m2)⋅(y−m1b2−b1m2m1−m2)+c=0⋯
=y2+m1m2x2+[(m1+m2)(m1b2−b1m2)−2m1m2(b2−b1)m1−m2⏟1.]x−(m1+m2)xy+[(m1+m2)(b2−b1)−2(m1b2−b1m2)m1−m2⏟2.]y+(m1b2−b1m2)2+m1m2(b2−b1)2−(m1+m2)(b2−b1)(m1b2−b1m2)(m1−m2)2⏟3.+c=0
calculate 1. :
(m1+m2)(m1b2−b1m2)−2m1m2(b2−b1)m1−m2=(m1+m2)m1b2−(m1+m2)b1m−2m1m2b2+2m1m2b1m1−m2⋯=−m1m2b2+m1m2b1+m21b2−m22b1m1−m2=b1m2(m1−m2)+b2m1(m1−m2)m1−m2=b1m2+b2m1
calculate 2. :
(m1+m2)(b2−b1)−2(m1b2−b1m2)m1−m2=m1b2−m1b1+m2b2−m2b1−2m1b2+2b1m2m1−m2=−m1b2+b1m2−m1b1+m2b2m1−m2=−b2(m1−m2)−b1(m1−m2)m1−m2=−b2−b1=−(b1+b2)
calculate 3. :
(m1b2−b1m2)2+m1m2(b2−b1)2−(m1+m2)(b2−b1)(m1b2−b1m2)(m1−m2)2⋯=−2m1m2b1b2+m21b1b2+m22b1b2(m1−m2)2=b1b2(m21−2m1m2+m22)(m1−m2)2=b1b2(m1−m2)2)(m1−m2)2=b1b2
So we have the same:
y2+m1m2x2+(m1b2+m2b1)x−(m1+m2)xy−(b1+b2)y+b1b2+c=0
