Question

Hi! I really need help please. I'm really desperate to pass our calculus but I'm having a hard time. If it is okay, can someone answer this question pls.

Find the coordinates of the center, vertices, foci, and extremities of the minor axis, the eccentricity and the equations of the directrices of the ellipse. 🙏

Answer 1

Having 20x²+12y-20x+12y=52

Answer 2

Sorry. Typo.

This is the final -> 20x²+12y²-20x+12y=52

Answer 3

first, the equation. This really should have been the easy part.
20x^2+12y^2-20x+12y=52
let's divide through by 4.
5x^2-5x + 3y^2+3y = 13
5(x^2 - x + 1/4) + 3(y^2 + y + 1/4) = 13 + 5/4 + 3/4
5(x - 1/2)^2 + 3(y - 1/2)^2 = 15
(x - 1/2)^2/3 + (y - 1/2)^2/5 = 1
so now we know that
a^2 = 5
b^2 = 3
c^2 = 2
and the major axis is vertical.
See what you can do with that. Confirm your results at

https://www.wolframalpha.com/input/?i=ellipse+20x%5E2%2B12y%5E2-20x%2B12y%3D52

Answer 4

Thank u. 😭🙏

Answer 5

dividing by 4 ... 5x² + 3y² - 5x + 3y = 13

rearranging ... 5(x² - x) + 3(y² + y) = 13

completing the squares ... 5(x - 1/2)^2 - 5/4 + 3(y + 1/2)^2 - 3/4 = 13
... 5(x - 1/2)^2 + 3(y + 1/2)^2 = 13 + 5/4 + 3/4 = 15

dividing by 15 ... [(x - 1/2)^2 / 3] + [(y + 1/2)^2 / 5] = 1

this is the equation of an ellipse centered at ... (1/2,-1/2)

the semi-major axis is vertical and equal to ... √5

the semi-minor axis is equal to ... √3

Answer 6

Thank u also, R_scott!