What is the vertices 16x^2+25y^2+32x-384=0 ?

The vertices 16x^2+25y^2+32x-384=0 is (4,0),(-6,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 16x^2+25y^2+32x-384=0

vertices

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

(4, 0), (- 6, 0)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} + 32 x - 384 = 0 :

Ellipse with center

(h, k) = (- 1, 0), a = 5, b = 4

(- 1 + 5, 0), (- 1 - 5, 0)

Refine

(4, 0), (- 6, 0)

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

25 x^{2} + 16 y^{2} + 100 x - 96 y - 156 = 0

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

100 x^{2} + 36 y^{2} = 225

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

What is the vertices 16x^2+25y^2+32x-384=0 ?
The vertices 16x^2+25y^2+32x-384=0 is (4,0),(-6,0)