What is the vertices (x^2)/(36)-((y-10)^2)/(64)=1 ?

The vertices (x^2)/(36)-((y-10)^2)/(64)=1 is (6,10),(-6,10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2)/(36)-((y-10)^2)/(64)=1

vertices

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1

(6, 10), (- 6, 10)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1 :

Right-left Hyperbola with

(h, k) = (0, 10), a = 6, b = 8

(0 + 6, 10), (0 - 6, 10)

Refine

(6, 10), (- 6, 10)

4 x^{2} - y^{2} + 32 x + 10 y + 38 = 0

4 x^{2} - y^{2} + 12 x + 5 = 0

\frac{y ^{2}}{36} - \frac{x ^{2}}{4} = 1

\frac{( x ^{2} )}{36} - \frac{( y ^{2} )}{25} = 1

16 x^{2} - y^{2} - 64 x - 2 y + 47 = 0

What is the vertices (x^2)/(36)-((y-10)^2)/(64)=1 ?
The vertices (x^2)/(36)-((y-10)^2)/(64)=1 is (6,10),(-6,10)