find the ellipse 4

Question

4.1 4x^2-45y^2= 180 4.2 9x^2 -16y^2-36x-32y-124+0

5. write the eq of the hyperbola w/ its center at the origin, tranverse axis on y-axis, eccentricity 2 sq. root of 3, length of the latus rectum 18.

6. determine what conic section is being represent by the ff.
a. y^2= 12x
b. 9x^2-16y^2+36x+32y+124=0
c. x^2+9y^2=9
d. x^2+y^2-10x-4y-21=0
e. 4y^2-12xy+10x^2+2x+1+0
f. 8x^2-4xy+5y^2-144=0

Answer 1

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)