What is the eccentricity (x^2)/(40)-(y^2)/(174)=1 ?

The eccentricity (x^2)/(40)-(y^2)/(174)=1 is (sqrt(107))/(2sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

eccentricity (x^2)/(40)-(y^2)/(174)=1

eccentricity

\frac{x ^{2}}{40} - \frac{y ^{2}}{174} = 1

\frac{107}{2 5}

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{40} - \frac{y ^{2}}{174} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 210, b = 174

= \frac{( 2 10 ) ^{2} + ( 174 ) ^{2}}{2 10}

Refine

= \frac{107}{2 5}

9 x^{2} - 4 y^{2} - 36 = 0

\frac{x ^{2}}{49} - \frac{y ^{2}}{16} = 1

4 y^{2} - 25 x^{2} = 100

16 x^{2} - 9 y^{2} + 96 x - 72 y + 144 = 0

\frac{( x + 5 ) ^{2}}{4} - \frac{( y - 2 ) ^{2}}{9} = 1

What is the eccentricity (x^2)/(40)-(y^2)/(174)=1 ?
The eccentricity (x^2)/(40)-(y^2)/(174)=1 is (sqrt(107))/(2sqrt(5))