Ellipses: Practice

The larger denominator, $25$, sits under $x^2$, so the major axis is horizontal and $a = 5$. The vertices are the ends of the major axis, so set $y = 0$.

$\frac{x^2}{25} = 1 \quad\Rightarrow\quad x = \pm 5$

The vertices are $(\pm 5, 0)$. The points $(0, \pm 4)$ are the co-vertices, the ends of the minor axis.

The three lengths form a right triangle with hypotenuse $a$, so $a^2 = b^2 + c^2$, which rearranges to $b^2 = a^2 - c^2$.

$b^2 = 13^2 - 5^2 = 169 - 25 = 144 \quad\Rightarrow\quad b = 12$

Adding instead of subtracting would give $\sqrt{194}$, and $a$ must be the largest of the three numbers.

By definition, that constant sum is the string length $2a$.

$2a = 20 \quad\Rightarrow\quad a = 10$

The sum equals the full major axis, $2a$, so halve it to get the semi-major axis $a$.

Here $a^2 = 100$ (the larger denominator, under $x^2$) and $b^2 = 64$, so the major axis is horizontal. Subtract to find $c$.

$c^2 = a^2 - b^2 = 100 - 64 = 36 \quad\Rightarrow\quad c = 6$

The foci sit on the major axis, $c$ units from the center, so they are $(\pm 6, 0)$. The points $(\pm 8, 0)$ use $b$ by mistake, and $\sqrt{164}$ comes from adding instead of subtracting.

The larger denominator is $36$, so $a^2 = 36$ and $a = 6$. The major axis runs from one vertex to the other, so it has length $2a$.

$2a = 2 \cdot 6 = 12$

The number $6$ is only the semi-major axis, the distance from the center to one vertex.

The larger denominator $49$ is under $x^2$, so the major axis is horizontal and the minor axis is vertical, with $b^2 = 9$. Set $x = 0$.

$\frac{y^2}{9} = 1 \quad\Rightarrow\quad y = \pm 3$

The co-vertices are $(0, \pm 3)$. The points $(\pm 7, 0)$ are the vertices.

The denominators are equal, so $a^2 = b^2 = 9$ and the focal distance collapses.

$c^2 = a^2 - b^2 = 9 - 9 = 0 \quad\Rightarrow\quad c = 0$

With $c = 0$ the two foci merge at the center, so there is no foci pair off-center and no long axis. Multiplying through by $9$ gives $x^2 + y^2 = 9$, a circle of radius $3$: the ellipse whose foci have collided.

Eccentricity compares how far the foci sit from the center with the size of the curve.

$e = \frac{c}{a}$

Because $0 \le c < a$ for every ellipse, this ratio always satisfies $0 \le e < 1$.

The derivation defines $b$ by $b^2 = a^2 - c^2$, which is legal precisely because $a > c$ makes the right side positive. Rearranged,

$a^2 = b^2 + c^2,$

which is the Pythagorean relation for the right triangle with legs $b$ and $c$ and hypotenuse $a$. A leg can never beat the hypotenuse, so $a \ge b$ and $a > c$ for every ellipse (and here $a > b$ is given outright). Any relation that makes $b$ or $c$ come out larger than $a$ is therefore wrong.

In standard position the foci are at $(\pm c, 0)$ when $a^2$ sits under $x^2$, so they are on the major axis, symmetric about the center.

$0 < c < a$

Because $c < a$, each focus is strictly inside the curve and strictly short of a vertex, and because $c > 0$ neither one sits on the minor axis.

Substitute each candidate and see which one makes the left side equal $1$. For $(4, 0)$:

$\frac{4^2}{16} + \frac{0^2}{4} = 1 + 0 = 1 \quad \checkmark$

The others fail: $(0, 4)$ gives $\frac{16}{4} = 4$, $(2, 2)$ gives $\frac{4}{16} + \frac{4}{4} = \frac{5}{4}$, and $(1, 2)$ gives $\frac{1}{16} + 1 = \frac{17}{16}$. The point $(4, 0)$ is the right vertex, since $a = 4$.

The vertices are on the $x$-axis, $6$ units from the center, so $a = 6$ and the major axis is horizontal. The co-vertices give $b = 4$.

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{x^2}{36} + \frac{y^2}{16} = 1$

The denominators are $a^2$ and $b^2$, not $a$ and $b$, so $6$ and $4$ must be squared.