What is the eccentricity (x^2)/(16)+(y-4)^2=1 ?

The eccentricity (x^2)/(16)+(y-4)^2=1 is (sqrt(15))/4

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eccentricity (x^2)/(16)+(y-4)^2=1

eccentricity

\frac{x ^{2}}{16} + (y - 4)^{2} = 1

\frac{15}{4}

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + (y - 4)^{2} = 1 :

Ellipse with center

(h, k) = (0, 4), a = 4, b = 1

= \frac{4 ^{2} - 1 ^{2}}{4}

\frac{4 ^{2} - 1 ^{2}}{4} = \frac{15}{4}

= \frac{15}{4}

9 x^{2} + 4 y^{2} - 18 x + 8 y - 23 = 0

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

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

\frac{( x - 3 ) ^{2}}{2} + \frac{( y - 4 ) ^{2}}{4} = 1

8 x^{2} + 4 y^{2} + 16 x - 12 y - 15 = 0

What is the eccentricity (x^2)/(16)+(y-4)^2=1 ?
The eccentricity (x^2)/(16)+(y-4)^2=1 is (sqrt(15))/4