What is the vertices 16x^2+36y^2-256x+448=0 ?

The vertices 16x^2+36y^2-256x+448=0 is (14,0),(2,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 16x^2+36y^2-256x+448=0

vertices

16 x^{2} + 36 y^{2} - 256 x + 448 = 0

(14, 0), (2, 0)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 36 y^{2} - 256 x + 448 = 0 :

Ellipse with center

(h, k) = (8, 0), a = 6, b = 4

(8 + 6, 0), (8 - 6, 0)

Refine

(14, 0), (2, 0)

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0

\frac{( x + 1 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{48} = 1

3 x^{2} + 4 y^{2} = 48

2 x^{2} + y^{2} - 98 = 0

3 x^{2} - 6 x + y^{2} + 4 y - 11 = 0

What is the vertices 16x^2+36y^2-256x+448=0 ?
The vertices 16x^2+36y^2-256x+448=0 is (14,0),(2,0)