What is the foci 16y^2+64y-x^2+2x=1 ?

The foci 16y^2+64y-x^2+2x=1 is (1,-2+2sqrt(17)),(1,-2-2sqrt(17))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci 16y^2+64y-x^2+2x=1

foci

16 y^{2} + 64 y - x^{2} + 2 x = 1

(1, - 2 + 217), (1, - 2 - 217)

Solution steps

(h, k + c), (h, k - c)

Calculate Hyperbola properties

16 y^{2} + 64 y - x^{2} + 2 x = 1 :

Up-down Hyperbola with

(h, k) = (1, - 2), a = 2, b = 8

(1, - 2 + c), (1, - 2 - c)

Compute

c :

c = 2^{2} + 8^{2} : 217

(1, - 2 + 217), (1, - 2 - 217)

- x^{2} + 9 y^{2} + 90 y + 189 = 0

\frac{( y - 3 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{45} = 1

36 x^{2} - 4 y^{2} - 216 x + 16 y + 452 = 0

2 y^{2} - \frac{x ^{2}}{2} = 2

\frac{y ^{2}}{25} - \frac{( x + 3 ) ^{2}}{56} = 1

What is the foci 16y^2+64y-x^2+2x=1 ?
The foci 16y^2+64y-x^2+2x=1 is (1,-2+2sqrt(17)),(1,-2-2sqrt(17))