HELP!!! find the equation,center,foci and etc. of the ellipse 3x^2+2y^2-36x-16y+134=0?

Question

find the equation, center, foci, major and minor axis, the latera recta, eccentricity, the directrix......

Answer 1

To answer that, you need to simplify the equation into the form
(x-h)^2/a^2 + (y-k)^2/b^2 = 1, so that you can immediately identify the center, foci, etc. And to do that, you can use the method of completing the squares.

3x^2+2y^2-36x-16y+134=0

(3x^2-36x)+(2y^2-16y)+134=0 [group related variables]

3(x^2-12x)+2(y^2-8y)+134=0 [factor out common factors]

3(x^2-12x+36-36)+2(y^2-8y+16-16)+134=0 [determine the third term for each expression. To do that, just take the half of the coefficient of the second term, then square. So, (12/2)^2 = 36, and (8/2)^2 = 16. Do not forget to subract it with the same number for equality]

3(x^2-12x+36)-108+2(y^2-8y+16)-32+134=0 [remove the unnecesary number out of each formed trinomial]

3(x-6)^2 + 2(y-4)^2 - 6=0 [get the perfect square binomials and simplify]

3(x-6)^2 + 2(y-4)^2 = 6

(x-6)^2/2 + (y-4)^2/3 = 1 [divide both sides by 6]

Therefore, h = 6, k = 4, a = sqrt(2) and b = sqrt(3). From these four numbers you can get all the properties of the ellipse that you wanted. For example, you can immediately see that the center is (h,k) = (6,4) and the semiminor axis has length sqrt(2) which runs horizontally, the semimajor axis has length sqrt(3) which runs vertically. The foci then are found directly above and below the center. To get the distance from the center to either focus, we use the formula
c = sqrt(b^2 - a^2)
c = sqrt(3 - 2)
c = 1.
Therefore the foci are
(h, k+c), (h, k-c)
(6, 5), (6, 3).

To get the length of the latus rectum, you use the formula:
L = 2a^2/b
L = 4/sqrt(3).

Answer 2

3(x^2-12x+36)-(36*3)+
2(y^2-8y+16)-(2*16)+134=0
3(x-6)^2+2(y-4)^2=6
(x-6)^2/2 + (y-4)^2/3=1--->equation
focus= sqrt(a^2-b^2)= √(9-4)=√5
foci: (0,√5) , (0, -√5)
centre= (6,4)
Length of Latus rectum= 2b^2/a= 2(2^2)/3=8/3
length of minor axis= 2b= 4
length of major axis =2a= 6
c=ae
e=c/a=√5/3
eq of directrices= +/- a/e= +/- 3/√5/3= +/- 9/√15

Answer 3

The given equation can be written as
3(x-6)^2+2(y-4)^2=4
using completing square.


ellipse equation:
(x-6)^2 /(4/3) +(y-4)^2 / 2 =1

a^2=4/3

b^2=2

center(6,4)