The eccentricity of the ellipse represented b | vedclass

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(A) Step $1$: Rewrite the equation by completing the square for $x$ and $y$ terms.$7(x^2 - 2x) + 16(y^2 + 4y) = 377$Step $2$: Add constants to complet

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$3/4$

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(A) Step $1$: Rewrite the equation by completing the square for $x$ and $y$ terms.$7(x^2 - 2x) + 16(y^2 + 4y) = 377$Step $2$: Add constants to complete the squares.$7(x^2 - 2x + 1) + 16(y^2 + 4y + 4) = 377 + 7(1) + 16(4)$$7(x - 1)^2 + 16(y + 2)^2 = 377 + 7 + 64 = 448$Step $3$: Divide by $448$ to get the standard form.$\frac{7(x - 1)^2}{448} + \frac{16(y + 2)^2}{448} = 1$$\frac{(x - 1)^2}{64} + \frac{(y + 2)^2}{28} = 1$Step $4$: Identify $a^2$ and $b^2$. Here $a^2 = 64$ and $b^2 = 28$.Step $5$: Calculate eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}}$.$e = \sqrt{1 - \frac{28}{64}} = \sqrt{\frac{64 - 28}{64}} = \sqrt{\frac{36}{64}} = \frac{6}{8} = \frac{3}{4}$.

The eccentricity of the ellipse represented by the equation $7x^2 + 16y^2 - 14x + 64y - 377 = 0$ is...

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