What is the eccentricity ((x-1)/4)^2+((y+2)/3)^2=1 ?

The eccentricity ((x-1)/4)^2+((y+2)/3)^2=1 is (sqrt(7))/4

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eccentricity ((x-1)/4)^2+((y+2)/3)^2=1

eccentricity

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

\frac{7}{4}

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (1, - 2), a = 4, b = 3

= \frac{4 ^{2} - 3 ^{2}}{4}

\frac{4 ^{2} - 3 ^{2}}{4} = \frac{7}{4}

= \frac{7}{4}

4 x^{2} + 9 y^{2} - 8 x - 36 y + 4 = 0

x^{2} + 4 y^{2} = 4

9 x^{2} + 36 y^{2} = 324

5 x^{2} + 7 y^{2} = 35

3 x^{2} + y^{2} + 18 x - 2 y - 8 = 0

What is the eccentricity ((x-1)/4)^2+((y+2)/3)^2=1 ?
The eccentricity ((x-1)/4)^2+((y+2)/3)^2=1 is (sqrt(7))/4