What is the vertices 16(x-1)^2+25(y-4)^2-400=0 ?

The vertices 16(x-1)^2+25(y-4)^2-400=0 is (6,4),(-4,4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 16(x-1)^2+25(y-4)^2-400=0

vertices

16 (x - 1)^{2} + 25 (y - 4)^{2} - 400 = 0

(6, 4), (- 4, 4)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

16 (x - 1)^{2} + 25 (y - 4)^{2} - 400 = 0 :

Ellipse with center

(h, k) = (1, 4), a = 5, b = 4

(1 + 5, 4), (1 - 5, 4)

Refine

(6, 4), (- 4, 4)

9 x^{2} + 4 y^{2} + 36 x + 24 y + 36 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{169} = 1

6 x^{2} + 9 y^{2} - 18 y - 45 = 0

7 x^{2} + 16 y^{2} - 112 = 0

9 x^{2} + 25 y^{2} + 90 x = 0

What is the vertices 16(x-1)^2+25(y-4)^2-400=0 ?
The vertices 16(x-1)^2+25(y-4)^2-400=0 is (6,4),(-4,4)