Ellipses: Practice

Here $a = 5$ and $b = 3$, so $c^2 = 25 - 9 = 16$ and $c = 4$. The chord is the vertical line $x = 4$, so substitute that value.

$\frac{16}{25} + \frac{y^2}{9} = 1 \quad\Rightarrow\quad \frac{y^2}{9} = \frac{9}{25} \quad\Rightarrow\quad y^2 = \frac{81}{25}$

So $y = \pm \frac{9}{5}$, and the chord runs from $\left(4, -\frac{9}{5}\right)$ to $\left(4, \frac{9}{5}\right)$.

$\text{length} = 2 \cdot \frac{9}{5} = \frac{18}{5}$

The value $6$ is the minor axis $2b$, which is the chord through the center, not through a focus.

Write the ellipse as $\frac{x^2}{A} + \frac{y^2}{B} = 1$ and substitute both points. Using $u = \frac{1}{A}$ and $v = \frac{1}{B}$ turns the two conditions into a linear system:

$4u + 3v = 1, \qquad 8u + 2v = 1$

Solve by elimination. Doubling $4u + 3v = 1$ gives $8u + 6v = 2$, and subtracting $8u + 2v = 1$ from it leaves $4v = 1$, so $v = \frac{1}{4}$ and $B = 4$. Back-substituting, $4u = 1 - \frac{3}{4} = \frac{1}{4}$, so $u = \frac{1}{16}$ and $A = 16$.

$\frac{x^2}{16} + \frac{y^2}{4} = 1$

Check the first point: $\frac{4}{16} + \frac{3}{4} = \frac{1}{4} + \frac{3}{4} = 1$.

Travelling from one focus out to a point $P$ and back to the other focus can never be shorter than going straight between the foci.

$PF_1 + PF_2 \ge F_1F_2 \quad\Longrightarrow\quad 2a \ge 2c$

Equality happens only when $P$ lies on the segment between the foci, and in that case the whole curve collapses to that segment. A genuine ellipse therefore needs the strict inequality $a > c$, which is also exactly what makes $b^2 = a^2 - c^2$ positive.

The base is the major axis, so $2a = 20$ and $a = 10$; the central height is the semi-minor axis, so $b = 6$. The arch is the upper half of $\frac{x^2}{100} + \frac{y^2}{36} = 1$. Substitute $x = 8$:

$\frac{64}{100} + \frac{y^2}{36} = 1 \quad\Rightarrow\quad \frac{y^2}{36} = 1 - \frac{16}{25} = \frac{9}{25}$

Multiply by $36$ and take the positive root:

$y^2 = \frac{324}{25} \quad\Rightarrow\quad y = \frac{18}{5} = 3.6$

The answer $4.8$ comes from assuming the height falls linearly, as if the arch were a straight ramp, which it is not.

The center is the midpoint of the foci, $(5, 2)$, and since the foci share a $y$-coordinate the major axis is horizontal with $c = 4$. Use the definition to get $a$: add the distances from $(5, 5)$ to the two foci.

$\sqrt{(5-1)^2 + (5-2)^2} + \sqrt{(5-9)^2 + (5-2)^2} = 5 + 5 = 10 = 2a$

So $a = 5$ and $b^2 = a^2 - c^2 = 25 - 16 = 9$, giving $\frac{(x - 5)^2}{25} + \frac{(y - 2)^2}{9} = 1$. The given point is the co-vertex, which is why each focal distance came out equal to $a$. Beware the option with denominators $16$ and $9$: the point $(5, 5)$ does satisfy it, but its foci are at $(5 \pm \sqrt{7}, 2)$, not the ones required.

The condition $2a = 2(2b)$ means $a = 2b$. Put that into the right triangle relation.

$c^2 = a^2 - b^2 = 4b^2 - b^2 = 3b^2 \quad\Rightarrow\quad c = b\sqrt{3}$

Now divide by $a = 2b$:

$e = \frac{c}{a} = \frac{b\sqrt{3}}{2b} = \frac{\sqrt{3}}{2} \approx 0.87$

The answer is independent of $b$, as it must be, because eccentricity measures shape and not size.

From the axis length, $2a = 16$, so $a = 8$ and $a^2 = 64$. From the eccentricity, $c = ea$.

$c = \frac{1}{2} \cdot 8 = 4$

Then $b^2 = a^2 - c^2 = 64 - 16 = 48$. The major axis lies along the $y$-axis, so $a^2 = 64$ goes under $y^2$ and $b^2 = 48$ under $x^2$.

$\frac{x^2}{48} + \frac{y^2}{64} = 1$

Substitute $x = 3$ into the equation.

$\frac{9}{9} + \frac{y^2}{4} = 1 \quad\Rightarrow\quad \frac{y^2}{4} = 0 \quad\Rightarrow\quad y = 0$

The only solution is $(3, 0)$, the right vertex. That makes sense: $a = 3$, so the ellipse never reaches beyond $x = 3$, and the line $x = 3$ just touches it there.

The two focal distances always add to the string length $2a$, which is the length of the major axis.

$r_1 + r_2 = 2a = 20$

So the other distance is $20 - 7 = 13$. No other information about the ellipse is needed, which is the whole point of the two-focus definition.

Write $c$ in terms of $a$ using $e = \frac{c}{a}$, so $c = \frac{4}{5}a$, and substitute into $b^2 = a^2 - c^2$.

$81 = a^2 - \frac{16}{25}a^2 = \frac{9}{25}a^2$

Multiply both sides by $\frac{25}{9}$:

$a^2 = 225 \quad\Rightarrow\quad a = 15$

Check it: $c = \frac{4}{5} \cdot 15 = 12$, and $b^2 = 225 - 144 = 81$, so $b = 9$ as required.

Every ellipse satisfies $a > c \ge 0$, so dividing by $a$ gives

$0 \le e = \frac{c}{a} < 1.$

The eccentricity can get as close to $1$ as you like, which flattens the curve, but it never reaches $1$: that would need $c = a$, and then $b^2 = a^2 - c^2 = 0$ and the curve would collapse to the segment between the foci. The other three statements are all true, since $e = 0$ forces $c = 0$, which merges the foci and makes $a = b$, a circle.

The vertices give $a = 10$, so the equation is $\frac{x^2}{100} + \frac{y^2}{b^2} = 1$. Substitute the point, using $\left(\frac{16}{5}\right)^2 = \frac{256}{25}$.

$\frac{36}{100} + \frac{256/25}{b^2} = 1 \quad\Rightarrow\quad \frac{256/25}{b^2} = \frac{16}{25}$

Cross-multiplying gives $16b^2 = 256$, so

$b^2 = 16 \quad\Rightarrow\quad b = 4.$

The value $\frac{16}{5}$ is only the height of the curve at $x = 6$, not the semi-minor axis, which is the height at $x = 0$.