3x2−4y2=z
x+2y−z=6
We will find z
z=x+2y−6
3x2−4y2=x+2y−6
4x2−3y2−12x−24y=−724(x2−3x)−3(y2+8y)=−72
4(x−3)2−3(y+2)2=−72+9−48=−111
1114(x−23)−1113(y+4)2=−14111(x−23)−3111(y+4)2=−1
This is a equation of conjugated hyperbola
center(23;−4)
semi−axes:a=4111,b=3111
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