What is the vertices 25x^2+4y^2-250x-16y+541=0 ?

The vertices 25x^2+4y^2-250x-16y+541=0 is (5,7),(5,-3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 25x^2+4y^2-250x-16y+541=0

vertices

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0

(5, 7), (5, - 3)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0 :

Ellipse with center

(h, k) = (5, 2), b = 5, a = 2

(5, 2 + 5), (5, 2 - 5)

Refine

(5, 7), (5, - 3)

3 x^{2} + 4 y^{2} = 12

\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{4} = 1

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

(x^{2} + 8 (y + 5)^{2}) = 20

What is the vertices 25x^2+4y^2-250x-16y+541=0 ?
The vertices 25x^2+4y^2-250x-16y+541=0 is (5,7),(5,-3)