Question

The eccentricity of the ellipse $\frac{\left(x-1\right)^{2}}{2} + \left(y + \frac{3}{4}\right)^{2} = \frac{1}{16}$ is

• A. $1 / \sqrt{2}$
• B. $1 / 2 \sqrt{2}$
• C. $1/2$
• D. $1/4$

Answer 1

Answer: (A)

$1 / \sqrt{2}$

Answer 2

We have, $\frac{(x-1)^{2}}{2}+(y+3 / 4)^{2}=\frac{1}{16}$ $\Rightarrow 8(x-1)^{2}+16(y+3 / 4)^{2}=1$ $\Rightarrow \frac{(x-1)^{2}}{1 / 8}+\frac{(y+3 / 4)^{2}}{1 / 16}=1$ $\therefore a^{2}=\frac{1}{8}$ and $b^{2}=\frac{1}{16}$ $\Rightarrow a=\frac{1}{2 \sqrt{2}}$ and $b=\frac{1}{4}$ $\therefore e =\sqrt{1-\frac{b^{2}}{a^{2}}} \,\,\,[\because a >b]$ $=\sqrt{1-\frac{1 / 16}{1 / 8}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$