The eccentricity of the ellipse (x - 3)^2 + ( | vedclass

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(B) Given equation: $(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$Multiply by $9$: $9(x - 3)^2 + 9(y^2 - 8y + 16) = y^2$$9(x - 3)^2 + 9y^2 - 72y + 144 = y^2$

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Given equation: $(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$Multiply by $9$: $9(x - 3)^2 + 9(y^2 - 8y + 16) = y^2$$9(x - 3)^2 + 9y^2 - 72y + 144 = y^2$$9(x - 3)^2 + 8y^2 - 72y + 144 = 0$$9(x - 3)^2 + 8(y^2 - 9y) = -144$$9(x - 3)^2 + 8(y - \frac{9}{2})^2 = -144 + 8(\frac{81}{4}) = -144 + 162 = 18$Divide by $18$: $\frac{(x - 3)^2}{2} + \frac{(y - 9/2)^2}{18/8} = 1$$\frac{(x - 3)^2}{2} + \frac{(y - 9/2)^2}{9/4} = 1$Here,$a^2 = 9/4$ and $b^2 = 2$ (since $a^2 > b^2$ is not true,let's re-evaluate: $a^2 = 2, b^2 = 9/4$ is incorrect,$a^2$ is the larger denominator).Since $9/4 = 2.25$ and $2 = 2$,$a^2 = 9/4$ and $b^2 = 2$.$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{2}{9/4} = 1 - \frac{8}{9} = \frac{1}{9}$$e = \frac{1}{3}$

The eccentricity of the ellipse $(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$ is

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