(4.1) 4x2-25y2=180, x2/45-5y2/36=1. a2=45, b2=36/5=7.2. The centre is (0,0), the origin.
Focal length c=√(a2+b2)=√(45+7.2)=√52.2=7.225 approx. (b) Foci at (±c,0)=(√52.2,0) and (-√52.2,0).
(c) Eccentricity e=c/a=√(52.2/45)=√(29/25)=√29/5=1.077 approx.
(a) Vertices lie along x-axis (as do foci) so y=0: 4x2=180, x=a=±√45=±3√5=6.708 approx. Vertices (-3√5,0) and (3√5,0).
(e) Asymptotes are the solution of 4x2-25y2=0, (2x)2=(5y)2, the lines 2x=5y, -2x=5y, which can be written:
y=⅖x and y=-⅖x.
(d) Latus rectum endpoints are found by solving for y when x=c (x coord of a focus).
4c2-25y2=180, 4×52.2-25y2=180, 25y2=208.8-180=144/5, 5y=12/√5=12√5/25=±1.0733 approx.
Endpoints of this LR=(√52.2,12√5/25) and (√52.2,-12√5/25). Length of LR=24√5/25=2.1466 approx. Because of symmetry the length of the other LR is the same.
(4.2) The hyperbola is not centralised because there are x and y terms. We need to find the centre.
The equation should be 9x2-16y2-36x-32y-124=0 (not "+0").
9x2-36x-(16y2+32y)=124, now, complete the squares:
9(x2-4x+4-4)-16(y2+2y+1-1)=124,
9(x-2)2-36-16(y+1)2+16=124,
9(x-2)2-16(y+1)2=124+36-16=144,
9(x-2)2/144-16(y+1)2/144=1,
(x-2)2/16-(y+1)2/9=1,
from which the centre is (2,-1), a2=16 (a=4) and b2=9, so c=√(16+9)=5; (c) e=c/a=5/4=1.25.
If we let X=x-2 and Y=y+1 we can find the other characteristics of the hyperbola more easily, by treating it as a centralised hyperbola using X and Y axes.
Equation of the hyperbola becomes 9X2-16Y2=144.
So, with reference to X and Y, the foci are at (±c,0), that is, X=±c, Y=0 and x-2=±c=±5, y+1=0, y=-1; x=2±5, making x=7 and -3 and the foci (7,-1) and (-3,-1).
Vertices lie along the X-axis so Y=0: 9X2=144, X2=16, X=±4, (X,Y)=(-4,0) and (4,0). x-2=±4, y+1=0, so (x,y) for the vertices is (2-4,-1) and (2+4,-1), that is, (a) vertices at (-2,-1) and (6,-1).
Asymptotes are found by solving 9X2-16Y2=0, 4Y=±3X becomes 4(y+1)=±3(x-2).
4y+4=3x-6 and 4y+4=-3x+6, from which (e) the asymptotes are 4y=3x-10 and 4y=-3x+2.
(d) LR endpoints are found by solving for Y when X=c=5 (x coord of a focus).
225-16Y2=144, 16Y2=225-144=81, Y=±9/4, so one LR has endpoints (5,9/4) and (5,-9/4) in the X-Y plane, making the length of the LR=9/2. (The length of the LR is unaffected by the positioning of the centre of the hyperbola. The endpoints of this LR in the x-y plane are (7,5/4) and (7,-13/4) because X=x-2-5, and Y=y+1=±9/4, so y=5/4 and -13/4.)