The eccentricity of the conic 4x^2 + 16y^2 - | vedclass

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(A) Given equation of the conic is $4x^2 + 16y^2 - 24x - 3y = 1$. Rearranging the terms: $4(x^2 - 6x) + 16(y^2 - \frac{3}{16}y) = 1$. Completing the s

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$\frac{\sqrt{3}}{2}$

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(A) Given equation of the conic is $4x^2 + 16y^2 - 24x - 3y = 1$. Rearranging the terms: $4(x^2 - 6x) + 16(y^2 - \frac{3}{16}y) = 1$. Completing the square: $4(x^2 - 6x + 9) + 16(y^2 - \frac{3}{16}y + \frac{9}{1024}) = 1 + 36 + \frac{9}{64}$. $4(x - 3)^2 + 16(y - \frac{3}{32})^2 = 37 + \frac{9}{64} = \frac{2377}{64}$. Dividing by the constant: $\frac{(x - 3)^2}{2377/256} + \frac{(y - 3/32)^2}{2377/1024} = 1$. Here,$a^2 = \frac{2377}{256}$ and $b^2 = \frac{2377}{1024}$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2377/1024}{2377/256}} = \sqrt{1 - \frac{256}{1024}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

The eccentricity of the conic $4x^2 + 16y^2 - 24x - 3y = 1$ is

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