4.1) 4x2-45y2=180 needs to be converted to standard form: x2/a2-y2/b2=±1. +1 means the hyperbolic curves are on the left and right of the vertical y-axis, while -1 means the hyperbolic curves are above and below the horizontal x-axis.
4x2-45y2=180, x2/45-y2/4=1, so a=√45=3√5, b=2, c2=a2+b2=45+4=49, c=7 (focal distance).
Eccentricity e=c/a or √(1+b2/a2)=√(1+4/45)=7/√45=7/(3√5) or 7√5/15 (rationalised).
Foci at (-7,0) and (7,0).
Centre (0,0). Length of LR=2b2/a=8/√45 or 8/(3√5)=8√5/15.
[When x=-7, 196-45y2=180, 45y2=16, y2=16/45, y=±4/√45 or ±4√5/15. So the endpoints of the left LR are at (-7,4√5/15) and (-7,-4√5/15) and at the right LR are at (7,4√5/15) and (7,-4√5/15), making the length of each LR=8√5/15.]
4.2) 9x2-16y2-36x-32y-124=0 has to be standardised to (x-h)2/a2-(y-k)2/b2=±1, where the centre of thre hyperbola is at (h,k). To achieve this, we need to complete two squares:
9x2-36x=9(x2-4x+4)-36=9(x-2)2-36;
16y2+32y=16(y2+2y+1)-16=16(y+1)2-16.
So 9x2-16y2-36x-32y-124=0 becomes 9(x-2)2-36-(16(y+1)2-16)-124=0,
9(x-2)2-36-16(y+1)2+16-124=0,
9(x-2)2-16(y+1)2=144,
(x-2)2/16-(y+1)2/9=1, a=4, b=3.
This hyperbola has its centre at (2,-1). This is the reference point for carrying out the other measurements. c2=a2+b2=25, so c=5 (focal length). The foci have to be located with respect to the centre. This puts the left focus at (2-5,-1)=(-3,-1) and the right focus at (2+5,-1)=(7,-1).
e=√(1+9/16)=√(25/16)=5/4.
The length of the LR is unaffected by the position of the centre=2b2/a=18/4=9/2.
5) Centre is (0,0)=(h,k). Transverse axis is y-axis so we have x2/a2-y2/b2=1, where a and b have yet to be found.
e=2√3=√(1+b2/a2); length of LR=18=2b2/a, so b2=9a, which can be substituted in e:
2√3=√(1+9a/a2)=√(1+9/a); 12=1+9/a after squaring; 11=9/a, a=9/11, a2=81/121, b2=81/11.
121x2/81-11y2/81=1, 121x2-11y2=81 is the equation of the hyperbola.
6a) parabola (sideways)
6b) hyperbola
6c) ellipse
6d) circle
6e) none (singularity at (-1,3/2)):
4y2-12xy+10x2+2x+1=0⇒(4y2-12xy+9x2)+(x2+2x+1)=0,
(2y-3x)2+(x+1)2=0, (2y-3x)2=-(x+1)2, which defines only one point (-1,3/2) since the RHS is negative except when x=-1, so no other points exist. All other values of x imply (2y-3x)2<0, for which there are no solutions.
6f) ellipse (tilted)