Ellipses: Practice

The center is $(3, -1)$. The larger denominator, $25$, sits under the $x$ term, so the major axis is horizontal and $a^2 = 25$, $b^2 = 16$.

$c^2 = 25 - 16 = 9 \quad\Rightarrow\quad c = 3$

The foci are $c = 3$ units from the center along the horizontal major axis, at $(3 \pm 3, -1)$, which is $(6, -1)$ and $(0, -1)$. Moving $3$ units vertically instead would put them on the minor axis, which never happens.

Standard form needs a right side of $1$, so divide every term by $225$.

$\frac{9x^2}{225} + \frac{25y^2}{225} = \frac{225}{225} \quad\Rightarrow\quad \frac{x^2}{25} + \frac{y^2}{9} = 1$

The coefficient $9$ on $x^2$ becomes the denominator $25$ under $x^2$, because $\frac{9}{225} = \frac{1}{25}$. Dividing turns the larger coefficient into the smaller denominator.

Both the focus and the vertex are on the $y$-axis, so the major axis is vertical, which means $a^2$ belongs under $y^2$. The vertex gives $a = 5$ and the focus gives $c = 3$.

$b^2 = a^2 - c^2 = 25 - 9 = 16$

So the equation is $\frac{x^2}{16} + \frac{y^2}{25} = 1$. Putting $9$ under $x^2$ would confuse $c^2$ with $b^2$.

Completing the square gives $(x - 3)^2 + 4(y + 1)^2 = 16$. Divide by $16$ to reach standard form:

$\frac{(x - 3)^2}{16} + \frac{(y + 1)^2}{4} = 1$

The larger denominator is $16$, so $a^2 = 16$ and $a = 4$. The major axis is the full segment between the vertices, so its length is $2a = 8$.

Use the right triangle relation $a^2 = b^2 + c^2$, in which $a$ is the hypotenuse.

$a^2 = 6^2 + 8^2 = 36 + 64 = 100 \quad\Rightarrow\quad a = 10$

Here $a$ must come out larger than both $b$ and $c$, and $10 > 8 > 6$ confirms it. Answers like $14 = b + c$ or $2 = c - b$ come from adding or subtracting the lengths themselves instead of their squares.

The ceiling is the top half of an ellipse. Its full length is the major axis, so $2a = 40$ and $a = 20$; its height at the center is the semi-minor axis, so $b = 12$.

$c^2 = a^2 - b^2 = 400 - 144 = 256 \quad\Rightarrow\quad c = 16$

Each focus sits $16$ meters from the center along the major axis. Adding instead of subtracting gives the impossible $\sqrt{544} > a$.

The constant sum is $2a$, so $2a = 10$ and $a = 5$. The foci are on the $x$-axis at distance $c = 2$ from the center, so the major axis is horizontal.

$b^2 = a^2 - c^2 = 25 - 4 = 21$

So the equation is $\frac{x^2}{25} + \frac{y^2}{21} = 1$. Using $2a = 10$ as $a$ would wrongly give the denominators $100$ and $96$.

The center is $(-2, 5)$. The larger denominator, $49$, sits under the $y$ term, so the major axis is vertical and $a = 7$.

$(-2, \; 5 \pm 7) = (-2, 12) \text{ and } (-2, -2)$

The vertices lie $a = 7$ units above and below the center. Moving $3$ units instead uses $b$, and moving horizontally uses the wrong axis.

The major axis belongs to the variable with the larger denominator, so a vertical major axis needs $9$ to beat $k$.

$9 > k \quad\text{and}\quad k > 0 \quad\Rightarrow\quad 0 < k < 9$

If $k > 9$ the major axis is horizontal instead. If $k = 9$ the denominators tie, $c = 0$, and the curve is a circle, which has no major axis at all, so $k = 9$ is excluded.

Halve the axis lengths: $a = 6$ and $b = 4$, so $a^2 = 36$ and $b^2 = 16$.

$\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 \quad \text{with } (h, k) = (-1, 4)$

Because the major axis is vertical, $a^2 = 36$ goes under the $y$ term. With $h = -1$, the factor is $x - (-1) = x + 1$, and with $k = 4$ it is $y - 4$.

Take the larger denominator minus the smaller one, since $c$ is a leg of the right triangle and $a$ is its hypotenuse.

$c^2 = 9 - 4 = 5 \quad\Rightarrow\quad c = \sqrt{5}$

Adding would give the impossible $\sqrt{13} > a = 3$, and forgetting the square root would give $5$.

First find $c$ from the denominators.

$c^2 = 49 - 40 = 9 \quad\Rightarrow\quad c = 3$

The foci are at $(\pm 3, 0)$, one on each side of the center, so the gap between them is $2c = 6$. The value $14$ is the major axis $2a$, not the focal separation.