Question

Find the center, vertices, and foci of the ellipse with equation 2x2 + 7y2 = 14

Answer 1

Answer on Question #53934 – Math – Analytic Geometry

Question

Find the center, vertices, and foci of the ellipse with equation $2x^{2} + 7y^{2} = 14$.

Solution

We transform the original equation in the next way: $\frac{x^2}{7} + \frac{y^2}{2} = 1 \Leftrightarrow \frac{x^2}{(\sqrt{7})^2} + \frac{y^2}{(\sqrt{2})^2} = 1$. So we have: $a = \sqrt{7}$, $b = \sqrt{2}$. Using the fact that canonical equation of ellipse has the form

$$\frac{(x - x_0)^2}{a^2} + \frac{(y - y_0)^2}{b^2} = 1$$

where $(x_0; y_0)$ is the center, we conclude that the ellipse

$\frac{x^2}{(\sqrt{7})^2} + \frac{y^2}{(\sqrt{2})^2} = 1$ has the center at the point $(0; 0)$. The vertices are

$A(-\sqrt{7}; 0), B(0; \sqrt{2}), C(\sqrt{7}; 0), D(0; -\sqrt{2})$. The focal length is equal to

$c = \sqrt{a^2 - b^2} = \sqrt{7 - 2} = \sqrt{5}$ so the foci are $F_2(-\sqrt{5}; 0)$ and $F_1(\sqrt{5}; 0)$.

**Answer:** Center: $(0; 0)$;

vertices: $(-\sqrt{7}; 0), (0; \sqrt{2}), (\sqrt{7}; 0), (0; -\sqrt{2})$;

foci: $(-\sqrt{5}; 0)$ and $(\sqrt{5}; 0)$.

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