What is the eccentricity (x^2)/(25)-(y^2)/(36)=1 ?

The eccentricity (x^2)/(25)-(y^2)/(36)=1 is (sqrt(61))/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

eccentricity (x^2)/(25)-(y^2)/(36)=1

eccentricity

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

\frac{61}{5}

Solution steps

= \frac{a ^{2} + b ^{2}}{a}

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 6

= \frac{5 ^{2} + 6 ^{2}}{5}

Refine

= \frac{61}{5}

x^{2} - 4 y^{2} - 4 x + 8 y - 16 = 0

\frac{( x - 2 ) ^{2}}{5} - \frac{( y + 1 ) ^{2}}{9} = 1

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

x^{2} - y^{2} = 25

\frac{x ^{2}}{45} - \frac{y ^{2}}{216} = 1

What is the eccentricity (x^2)/(25)-(y^2)/(36)=1 ?
The eccentricity (x^2)/(25)-(y^2)/(36)=1 is (sqrt(61))/5