Q.Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(0, \pm \sqrt{5})$, ends of minor axis $(\pm 1, 0)$.

The major axis lies along the $y$-axis (since the ends are at $(0, \pm\sqrt{5})$) and the minor axis along the $x$-axis (ends at $(\pm 1, 0)$). This tells us the ellipse has a vertical major axis.

For an ellipse centered at the origin with vertical major axis, the standard form is:

where $a$ is the semi-major axis and $b$ is the semi-minor axis.

From the given ends:

• Semi-major axis: $a = \sqrt{5}$, so $a^2 = 5$
• Semi-minor axis: $b = 1$, so $b^2 = 1$ …

An ellipse is the locus of all points for which the sum of distances to two fixed points (the foci) is constant. The major axis is the longest diameter, and the minor axis is the shortest. The standard equation depends on which axis is major.

When the major axis is horizontal, the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$.

When the major axis is vertical, the equation is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ with $a > b$.

Here $a$ is always the semi-major axis length and $b$ the semi-minor axis length.

The problem tells us the ends of the major axis are at $(0, \pm\sqrt{5})$ and the ends of the minor axis are at $(\pm 1, 0)$. This immediately reveals the orientation.

1. Identify the orientation.

   The major axis endpoints lie on the $y$-axis, so the major axis is vertical. The minor axis endpoints lie on the $x$-axis, so the minor axis is horizontal.

2. Find the semi-major axis length $a$.

   The endpoints of the major axis are $(0, \pm\sqrt{5})$, which means the ellipse extends from $y = -\sqrt{5}$ to $y = +\sqrt{5}$. The distance from the center (origin) to either endpoint is $a = \sqrt{5}$.

3. Find the semi-minor axis length $b$.

   The endpoints of the minor axis are $(\pm 1, 0)$, so the ellipse extends from $x = -1$ to $x = +1$. The distance from the center to either endpoint is $b = 1$. …