Ellipses: Additional Practice (Set 2)

Since $k < 25$, the larger denominator is $25$, so $a^2 = 25$, $a = 5$, and $b^2 = k$. Turn the eccentricity into a value of $c$:

$e = \frac{c}{a} = \frac{3}{5} \quad\Rightarrow\quad c = 3$

Now use $b^2 = a^2 - c^2$:

$k = b^2 = 25 - 9 = 16$

The restriction $0 < k < 25$ matters. Without it, a value of $k$ larger than $25$ would make $y$ the major variable and produce a second, different answer, so the problem would have two solutions.

The squared terms vanish at $x = -4$ and $y = 3$, so the center is $(-4, 3)$. The larger denominator, $100$, sits under the $x$ term, so the major axis is horizontal with $a = 10$ and $b = 8$.

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

The foci are $6$ units left and right of the center, at $(-4 \pm 6, 3)$, which is $(-10, 3)$ and $(2, 3)$. Using $c = 8$ would be using $b$ by mistake.

Substitute $y = x + 2$ into the ellipse and clear denominators by multiplying through by $16$.

$x^2 + 4(x + 2)^2 = 16 \quad\Rightarrow\quad x^2 + 4x^2 + 16x + 16 = 16$

That simplifies to $5x^2 + 16x = 0$, so $x(5x + 16) = 0$ and $x = 0$ or $x = -\frac{16}{5}$.

$x = 0 \Rightarrow y = 2, \qquad x = -\frac{16}{5} \Rightarrow y = -\frac{16}{5} + 2 = -\frac{6}{5}$

The point $(-2, 0)$ is on the line but not on the ellipse, since $\frac{4}{16} = \frac{1}{4} \ne 1$.

Group the terms and factor out the leading coefficients:

$16\left(x^2 + 2x\right) + 25\left(y^2 - 6y\right) = 159$

Complete each square. Adding $1$ inside the first parentheses adds $16 \cdot 1 = 16$ to the left; adding $9$ inside the second adds $25 \cdot 9 = 225$. Add both to the right:

$16(x + 1)^2 + 25(y - 3)^2 = 159 + 16 + 225 = 400$

Divide by $400$, so that $\frac{16}{400} = \frac{1}{25}$ and $\frac{25}{400} = \frac{1}{16}$:

$\frac{(x + 1)^2}{25} + \frac{(y - 3)^2}{16} = 1$

Group and factor:

$25\left(x^2 - 4x\right) + 4\left(y^2 + 6y\right) = -36$

Complete each square. Adding $4$ inside the first parentheses adds $25 \cdot 4 = 100$ to the left, and adding $9$ inside the second adds $4 \cdot 9 = 36$, so add both amounts on the right:

$25(x - 2)^2 + 4(y + 3)^2 = -36 + 100 + 36 = 100$

Divide by $100$:

$\frac{(x - 2)^2}{4} + \frac{(y + 3)^2}{25} = 1$

The center is $(2, -3)$, and since $25 > 4$ the major axis is vertical.

The larger denominator gives $a^2 = 25$, so $a = 5$, and $b^2 = 16$.

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

Now apply the definition:

$e = \frac{c}{a} = \frac{3}{5}$

Shifting the center never changes $a$, $b$, or $c$, so the eccentricity does not depend on where the ellipse sits. The value $\frac{4}{5}$ is $\frac{b}{a}$.

The center is $(2, -3)$. The larger denominator, $25$, sits under the $y$ term, so the major axis is vertical, $a^2 = 25$ and $b^2 = 4$.

$c^2 = 25 - 4 = 21 \quad\Rightarrow\quad c = \sqrt{21}$

The foci lie $\sqrt{21}$ units above and below the center, at $\left(2, -3 \pm \sqrt{21}\right)$. The points $(2, -3 \pm 5)$ are the vertices, and $\sqrt{29}$ comes from adding the denominators instead of subtracting.

The foci are on the $y$-axis, so the major axis is vertical and $a^2$ belongs under $y^2$, with $c = 6$. The minor axis has length $2b = 16$, so $b = 8$.

$a^2 = b^2 + c^2 = 64 + 36 = 100$

So the equation is $\frac{x^2}{64} + \frac{y^2}{100} = 1$, with $a = 10 > b = 8$ as it must be. Placing $100$ under $x^2$ would put the foci on the wrong axis.

With $a$ fixed, $c = ea$, so a smaller $e$ means the foci sit closer to the center. That directly controls the width:

$b = a\sqrt{1 - e^2}$

For ellipse A, $b = a\sqrt{1 - 0.09} \approx 0.95a$, nearly as wide as it is long. For ellipse B, $b = a\sqrt{1 - 0.49} \approx 0.71a$, noticeably flatter. Since a circle is the case $e = 0$, the ellipse with the smaller eccentricity is the rounder one.

The larger denominator gives $a = 13$, so the two focal distances add to $2a = 26$.

$r_2 = 2a - r_1 = 26 - 17 = 9$

The answer is consistent: $c = \sqrt{169 - 25} = 12$, so focal distances must lie between $a - c = 1$ and $a + c = 25$, and both $17$ and $9$ fall inside that range.

The gap between the foci is $2c$ and the constant sum is $2a$.

$2c = 6 \Rightarrow c = 3, \qquad 2a = 10 \Rightarrow a = 5$

Now find $b$ from the right triangle relation:

$b^2 = a^2 - c^2 = 25 - 9 = 16 \quad\Rightarrow\quad b = 4$

The minor axis is the full segment between the co-vertices, so it has length $2b = 8$.

Here $a = 13$ and $b = 5$, so $c^2 = 169 - 25 = 144$ and $c = 12$. The chord is the vertical line $x = 12$, so substitute it:

$\frac{144}{169} + \frac{y^2}{25} = 1 \quad\Rightarrow\quad \frac{y^2}{25} = \frac{25}{169}$

So $y^2 = \frac{625}{169}$ and $y = \pm \frac{25}{13}$. The chord runs from one of those points to the other:

$\text{length} = 2 \cdot \frac{25}{13} = \frac{50}{13}$

The value $10$ is the minor axis, the chord through the center rather than through a focus.