What is the eccentricity ((x-4)^2)/5+((y-2)^2)/4 =1 ?

The eccentricity ((x-4)^2)/5+((y-2)^2)/4 =1 is 1/(sqrt(5))

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eccentricity ((x-4)^2)/5+((y-2)^2)/4 =1

eccentricity

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

\frac{1}{5}

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (4, 2), a = 5, b = 2

= \frac{( 5 ) ^{2} - 2 ^{2}}{5}

\frac{( 5 ) ^{2} - 2 ^{2}}{5} = \frac{1}{5}

= \frac{1}{5}

49 x^{2} - 98 x + 64 y^{2} + 256 y - 2831 = 0

\frac{( x ^{2} )}{16} + \frac{( y ^{2} )}{25} = 1

\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1

4 x^{2} + 32 x + 25 y^{2} - 250 y + 589 = 0

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1

What is the eccentricity ((x-4)^2)/5+((y-2)^2)/4 =1 ?
The eccentricity ((x-4)^2)/5+((y-2)^2)/4 =1 is 1/(sqrt(5))