Ellipses & Hyperbolas - Algebra 3

1. A hyperbola has vertices $(1,3)$ and $(7,3)$, and one focus is at $(9,3)$. What are its asymptotes?
   Answer: $y-3=\pm \frac{4}{3}(x-4)$ The center is $(4,3)$, and $a=3$. The focus is 5 units from the center, so $c=5$. Then $b^2=25-9=16$, so $b=4$. For a horizontal hyperbola, the asymptote slope is $b/a=4/3$, giving $y-3=\pm \frac{4}{3}(x-4)$.
2. Which equation represents a hyperbola with center at the origin, focus $(0,\sqrt{34})$, and asymptote $y=\frac{3}{5}x$?
   Answer: $\frac{y^2}{9}-\frac{x^2}{25}=1$ The focus lies on the $y$-axis, so the vertical form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ is required. Its asymptote slope is $a/b$. Setting $a/b=3/5$ and $c=\sqrt{34}$, the values $a=3$, $b=5$ work because $3^2+5^2=34$. Hence the equation is $\frac{y^2}{9}-\frac{x^2}{25}=1$.
3. For which values of $k$ does $\frac{x^2}{k-4}+\frac{y^2}{8-k}=1$ represent a hyperbola with vertices on the $y$-axis?
   Answer: $k<4$ Vertices on the $y$-axis mean the positive term is the $y$-term, so we need $8-k>0$ and $k-4<0$. Both conditions hold exactly when $k<4$. If $k>8$, the $x$-term is positive and the hyperbola is horizontal; if $4<k<8$, both denominators are positive and the graph is an ellipse.
4. An ellipse has center $(2,-3)$, one vertex at $(7,-3)$, and one focus at $(5,-3)$. Which equation represents the ellipse?
   Answer: $\frac{(x-2)^2}{25}+\frac{(y+3)^2}{16}=1$ The vertex is 5 units from the center, so $a=5$. The focus is 3 units from the center, so $c=3$. Thus $b^2=25-9=16$. Since both the vertex and focus lie on the horizontal line $y=-3$, the major axis is horizontal, giving $\frac{(x-2)^2}{25}+\frac{(y+3)^2}{16}=1$.
5. An ellipse is centered at the origin. Its vertices are $(0,6)$ and $(0,-6)$, and one focus is $(0,\sqrt{11})$. Which equation represents the ellipse?
   Answer: $\frac{x^2}{25}+\frac{y^2}{36}=1$ Whenever you see an ellipse centered at the origin, first locate the vertices: they tell you the major axis and the value of $a$. Since the vertices are $(0,6)$ and $(0,-6)$, the major axis is vertical, so $a=6$, and $a^2=36$. The focus $(0,\sqrt{11})$ gives $c=\sqrt{11}$, so $c^2=11$. For an ellipse, $b^2=a^2-c^2=36-11=25$. Because the major axis is vertical, the larger denominator belongs under $y^2$: $\frac{x^2}{25}+\frac{y^2}{36}=1$. The choice $\frac{x^2}{36}+\frac{y^2}{25}=1$ swaps the denominators, treating the major axis as horizontal. The choice $\frac{x^2}{11}+\frac{y^2}{36}=1$ puts $c^2=11$ under $x^2$, confusing the focal distance with the semi-minor axis. The choice $\frac{x^2}{25}+\frac{y^2}{11}=1$ puts $c^2=11$ under $y^2$, losing the fact that the $y$-direction reaches to the vertices. Only the equation with $36$ under $y^2$ and $25$ under $x^2$ matches the vertical ellipse. Study tip: always identify whether the major axis is vertical or horizontal first, then use $b^2=a^2-c^2$. That prevents the most common mistake on this exam: swapping the denominators or using $c^2$ as $b^2$.
6. Consider the ellipse $9x^2+25y^2-36x+50y-164=0$. What are the endpoints of its major axis?
   Answer: $(7,-1)$ and $(-3,-1)$ Whenever you see a general quadratic like this, your first move should be completing the square to put the ellipse in standard form. That reveals the center and which axis is major. Group the $x$ and $y$ terms: $$9(x^2-4x)+25(y^2+2y)=164.$$ Complete the square: $$9[(x-2)^2-4]+25[(y+1)^2-1]=164$$ $$9(x-2)^2+25(y+1)^2=225.$$ Divide by 225: $$\frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1.$$ The center is $(2,-1)$. Since $25>9$, the major axis is horizontal, with $a=5$. So the endpoints are $(2\pm5,-1)$, which are $(7,-1)$ and $(-3,-1)$. The choices $(2,2)$ and $(2,-4)$ are the endpoints of the minor axis, because they use $b=3$ vertically — a classic swap-the-axes error. The choices $(6,-1)$ and $(-2,-1)$ are actually the foci, since $c=\sqrt{25-9}=4$, not the vertices. Finally, $(-7,1)$ and $(3,1)$ reverse both the center and the signs: they come from treating the center as $(-2,1)$ instead of $(2,-1)$. Your study tip: always write the standard form before finding endpoints. The larger denominator tells you the major axis direction; $a^2$ is that denominator, not the distance from the center to the focus. Check that your endpoints are exactly $a$ units from the center along that axis.
7. The conic $4x^2-9y^2+8x+36y-16=0$ is a hyperbola. What are its vertices?
   Answer: The vertices are $\left(-1,\frac{10}{3}\right)$ and $\left(-1,\frac{2}{3}\right)$. When you see a general second-degree conic like this, completing the square reveals both its center and orientation. Group terms: $$4(x^2+2x)-9(y^2-4y)-16=0.$$ Complete the squares: $$4[(x+1)^2-1]-9[(y-2)^2-4]-16=0,$$ which simplifies to $$9(y-2)^2-4(x+1)^2=16.$$ Divide by $16$: $$\frac{(y-2)^2}{16/9}-\frac{(x+1)^2}{4}=1.$$ This is a vertical hyperbola centered at $(-1,2)$, with $a^2=\frac{16}{9}$, so $a=\frac43$. Therefore the vertices are $\frac43$ units above and below the center, along $x=-1$: $$\left(-1,2+\frac43\right)=\left(-1,\frac{10}{3}\right),\qquad \left(-1,2-\frac43\right)=\left(-1,\frac23\right).$$ The choices with $x=1$, such as $\left(1,\frac{10}{3}\right)$ and $\left(1,\frac23\right)$, use the wrong center sign from $(x+1)^2$. The choice $\left(-1,2\pm\frac{2\sqrt{13}}3\right)$ gives the foci, since $c=\sqrt{a^2+b^2}=\sqrt{\frac{16}{9}+4}=\frac{2\sqrt{13}}3$, not the vertices. The choice $(-1,4)$ and $(-1,0)$ uses distance $2$ from the center, which is $b$, not $a$; the relevant denominator is $\frac{16}{9}$, not $4$. Study tip: after writing a hyperbola in standard form, look at which squared term is positive — that term gives the transverse axis and $a^2$. Vertices are center $\pm a$ along that axis; foci come from $c^2=a^2+b^2$.
8. An elliptical arch is modeled by the upper half of $\frac{x^2}{400}+\frac{y^2}{256}=1$, with distances in feet. What is the height of the arch 8 ft from its center?
   Answer: $\frac{16\sqrt{21}}{5}$ Set $x=8$: $\frac{64}{400}+\frac{y^2}{256}=1$, so $\frac{y^2}{256}=\frac{336}{400}=\frac{21}{25}$. Taking the positive square root gives $y=\sqrt{256 \cdot \frac{21}{25}}=\frac{16\sqrt{21}}{5}$.
9. An ellipse has endpoints of the minor axis at $(-5,3)$ and $(1,3)$, and one focus at $(-2,5)$. Which equation represents the ellipse?
   Answer: $\frac{(x+2)^2}{9}+\frac{(y-3)^2}{13}=1$ The minor-axis endpoints have midpoint $(-2,3)$, so the horizontal semi-axis is $b=3$. The focus $(-2,5)$ lies on the vertical major axis, so $c=2$. For an ellipse, $a^2=b^2+c^2=9+4=13$. Since the major axis is vertical, the larger denominator goes under the $y$-term: $\frac{(x+2)^2}{9}+\frac{(y-3)^2}{13}=1$.
10. An ellipse has foci $(\pm 6,0)$ and passes through $(3,2\sqrt{10})$. Which equation represents the ellipse?
    Answer: $\frac{x^2}{81}+\frac{y^2}{45}=1$ For a horizontal ellipse, $c=6$ and $b^2=a^2-36$. Substituting $(3,2\sqrt{10})$ gives $\frac{9}{a^2}+\frac{40}{a^2-36}=1$, which simplifies to $a^4-85a^2+324=0$. The valid root is $a^2=81$, since $a^2=4$ is less than $c^2=36$. Thus $b^2=81-36=45$, so the equation is $\frac{x^2}{81}+\frac{y^2}{45}=1$. The choice $\frac{x^2}{63}+\frac{y^2}{27}=1$ has the correct focal distance, but the given point does not lie on it.
11. How many of the following equations have graphs that are hyperbolas: $\frac{x^2}{16}-\frac{y^2}{9}=1$, $\frac{x^2}{16}+\frac{y^2}{9}=-1$, $-\frac{x^2}{16}+\frac{y^2}{9}=1$, and $\frac{x^2}{16}+\frac{y^2}{9}=1$?
    Answer: 2 The first equation is a horizontal hyperbola. The second has no real graph, since a sum of nonnegative terms cannot equal $-1$. The third is $\frac{y^2}{9}-\frac{x^2}{16}=1$, a vertical hyperbola. The fourth is an ellipse. Therefore exactly 2 of the equations are hyperbolas.
12. Consider the conic $9x^2 - 4y^2 - 54x - 16y - 79 = 0$. Which statement about its graph is true?
    Answer: The vertices are $(-1,-2)$ and $(7,-2)$. Completing the square gives $9(x-3)^2 - 4(y+2)^2 = 144$, so $\frac{(x-3)^2}{16} - \frac{(y+2)^2}{36}=1$. The center is $(3,-2)$, $a=4$, and $b=6$. The transverse axis is horizontal, so the vertices are $(3 \pm 4, -2)$, namely $(-1,-2)$ and $(7,-2)$. The focal distance is $c=\sqrt{52}=2\sqrt{13}$, so the foci are $(3 \pm 2\sqrt{13}, -2)$, not vertically displaced. The asymptote slope is $6/4=3/2$, not $2/3$, and the center is $(3,-2)$, not $(3,2)$.
13. A conic with center at the origin has vertices $(\pm 4,0)$ and passes through $(2,3)$. Which point is a focus of this conic?
    Answer: $(2,0)$ If it were a hyperbola, $\frac{x^2}{16}-\frac{y^2}{b^2}=1$ would require $\frac{1}{4}-\frac{9}{b^2}=1$, which is impossible. Thus it is an ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$. Substituting $(2,3)$ gives $\frac{1}{4}+\frac{9}{b^2}=1$, so $b^2=12$. Then $c^2=16-12=4$, so the foci are $(\pm 2,0)$; $(2,0)$ is one of them.
14. A hyperbola has center $(-1,5)$, vertex $(-4,5)$, and focus $(-7,5)$. Which equation represents one of its asymptotes?
    Answer: $y-5=\pm \sqrt{3}(x+1)$ The vertex is 3 units from the center, so $a=3$. The focus is 6 units from the center, so $c=6$. Therefore $b^2=36-9=27$, so $b=3\sqrt{3}$. The transverse axis is horizontal, so the asymptote slope is $b/a=\sqrt{3}$. Thus one asymptote pair is $y-5=\pm \sqrt{3}(x+1)$.
15. Which statement is false about the graphs of $\frac{x^2}{16}+\frac{y^2}{9}=1$ and $\frac{x^2}{16}-\frac{y^2}{9}=1$?
    Answer: They have the same $y$-intercepts. The ellipse has $y$-intercepts $(0,\pm 3)$, but substituting $x=0$ into the hyperbola gives $-\frac{y^2}{9}=1$, which has no real solution. Thus the hyperbola has no $y$-intercepts. Both graphs have $x$-intercepts $(\pm 4,0)$, both are centered at the origin, and only the hyperbola has asymptotes.
16. A hyperbola centered at $(-2,4)$ has a vertex at $(-2,7)$ and a focus at $(-2,9)$. Which equation represents the hyperbola?
    Answer: $\frac{(y-4)^2}{9}-\frac{(x+2)^2}{16}=1$ Whenever you see a hyperbola question, start by locating the center and deciding whether the transverse axis is horizontal or vertical. Here the center is $(-2,4)$, and both the vertex $(-2,7)$ and focus $(-2,9)$ lie directly above it, so the hyperbola opens vertically — the $y$-term comes first. The distance from the center to the vertex is $a=3$. The distance from the center to the focus is $c=5$. For a hyperbola, $c^2=a^2+b^2$, so $25=9+b^2$, giving $b^2=16$. The vertical standard form is $\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$. Substituting $h=-2$, $k=4$ gives $\frac{(y-4)^2}{9}-\frac{(x+2)^2}{16}=1$, the correct equation. The choice $\frac{(x+2)^2}{16}-\frac{(y-4)^2}{9}=1$ would open horizontally, but your vertices and focus are above/below the center, not left/right. It also uses the denominators in the wrong positions. The option $\frac{(y-4)^2}{9}-\frac{(x-2)^2}{16}=1$ has the right vertical orientation butn $x-2$ shifts the center to $(2,4)$, when you need $x+2$ for center $x=-2$. The choice $\frac{(y+4)^2}{9}-\frac{(x-2)^2}{16}=1$ shifts both coordinates, placing the center at $(2,-4)$, which is incorrect. A quick check: from center, the vertex and focus both move in the $y$-direction, so vertical form. Then the signs in parentheses must match the center: $x+2$ and $y-4$. Memorize this pattern — orientation decides which term is positive, and the parentheses reveal the center.
17. An elliptical archway is modeled by the upper half of an ellipse centered at the origin. The archway is 20 feet wide at its base and 6 feet high at the center. Which equation describes the full ellipse?
    Answer: $\frac{x^2}{100}+\frac{y^2}{36}=1$ Whenever you see an ellipse problem, start by recalling the standard form: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a$ is the horizontal semi-axis and $b$ is the vertical semi-axis. Here, the archway is the upper half of an ellipse centered at the origin, so its base lies on the $x$-axis. A width of 20 feet means the full ellipse crosses from $-10$ to $10$ on the $x$-axis, so $a=10$ and $a^2=100$. The height of 6 feet at the center means the top of the arch reaches $y=6$, so $b=6$ and $b^2=36$. Therefore the full ellipse is $\frac{x^2}{100}+\frac{y^2}{36}=1.$ The choice $\frac{x^2}{36}+\frac{y^2}{100}=1$ swaps the axes, giving a 12-foot-wide, 20-foot-tall ellipse. The choice $\frac{x^2}{100}+\frac{y^2}{12}=1$ uses 12 instead of $6^2$, confusing the height with its square. The choice $\frac{x^2}{400}+\frac{y^2}{36}=1$ squares the full width 20 instead of the semi-width 10, which would make the ellipse 40 feet wide. On exam day, remember: the denominators in the ellipse equation are the squares of the semi-axes, not the full widths or heights. Always halve the total width before squaring it.
18. Which of the following is an equation of one asymptote of the hyperbola $\frac{(x-3)^2}{16}-\frac{(y+2)^2}{9}=1$?
    Answer: $y+2=\frac{3}{4}(x-3)$ When you see a hyperbola in standard form, your first move is to locate its center and decide whether it opens left/right or up/down. Here the equation is $\frac{(x-3)^2}{16}-\frac{(y+2)^2}{9}=1$, so the center is $(3,-2)$. Because the $x$-term is positive, the transverse axis is horizontal; thus the asymptote slopes are $\pm \frac{b}{a}$, where $a^2=16$ gives $a=4$ and $b^2=9$ gives $b=3$. Asymptotes pass through the center, so they have the form $y - (-2)=\pm \frac{3}{4}(x-3)$, or $y+2=\pm \frac{3}{4}(x-3)$. That makes $y+2=\frac{3}{4}(x-3)$ the correct asymptote. The choice $y-2=\frac{3}{4}(x+3)$ uses the wrong center: it treats the center as $(-3,2)$ instead of $(3,-2)$. The choice $y+2=\frac{4}{3}(x-3)$ has the right center but uses the reciprocal slope $\frac{a}{b}=\frac{4}{3}$, which would be the slope if the hyperbola opened vertically. The remaining choice $y-2=\frac{4}{3}(x+3)$ combines both errors: wrong center and reversed slope ratio. A reliable study tip: for any hyperbola, write the center from the shifted coordinates, identify which squared term is positive to know the direction, and then form asymptotes using $\frac{b}{a}$ for horizontal hyperbolas but $\frac{a}{b}$ for vertical ones. Always keep the slope paired with the actual center of the hyperbola.
19. A hyperbola has equation $\frac{y^2}{25}-\frac{x^2}{16}=1$. Which statement about its asymptotes is correct?
    Answer: They intersect at $(0,0)$ and have slopes $\pm\frac{5}{4}$. Whenever you see a hyperbola in standard form, first identify which squared term is positive. Here $\frac{y^2}{25}-\frac{x^2}{16}=1$ is a vertical hyperbola because $y^2$ comes first, so its center is $(0,0)$ and its asymptotes pass through the center with slopes $\pm \frac{a}{b}$, where $a^2=25$ and $b^2=16$. Thus $a=5$, $b=4$, giving slopes $\pm \frac{5}{4}$. The correct statement is the one saying they intersect at $(0,0)$ and have slopes $\pm\frac{5}{4}$. The choice claiming they intersect at $(5,4)$ confuses the center with a point on or related to the rectangle; the asymptotes always intersect at the hyperbola's center, not at $(a,b)$. The choice with slopes $\pm\frac{4}{5}$ uses $\frac{b}{a}$, which is the slope formula for a horizontal hyperbola like $\frac{x^2}{25}-\frac{y^2}{16}=1$, not this one. The choice with slopes $\pm\frac{25}{16}$ incorrectly squares the denominators instead of taking square roots to get $a$ and $b$. A reliable strategy: for hyperbolas, find the center from the numerators, then read $a^2$ and $b^2$ from the denominators. If $y^2$ is positive, slopes are $\pm\frac{a}{b}$; if $x^2$ is positive, slopes are $\pm\frac{b}{a}$. Also remember the center is always the asymptotes' intersection point.
20. A hyperbola is centered at $(-1,3)$, has vertices $(-4,3)$ and $(2,3)$, and its asymptotes have slopes $\pm\frac{4}{3}$. Which equation represents the hyperbola?
    Answer: $\frac{(x+1)^2}{9}-\frac{(y-3)^2}{16}=1$ Whenever you see a hyperbola, start by translating the center into parentheses. Since the center is $(-1,3)$, the equation must contain $(x+1)$ and $(y-3)$. The vertices $(-4,3)$ and $(2,3)$ lie horizontally from the center, so the hyperbola opens left/right: the $x$-term must be positive, and the distance from center to a vertex is $3$, so $a^2=9$. For a horizontal hyperbola, asymptote slopes are $\pm \frac{b}{a}$. Given $\pm\frac{4}{3}$ and $a=3$, you get $b=4$, so $b^2=16$. This yields $\frac{(x+1)^2}{9}-\frac{(y-3)^2}{16}=1$. The equation with $\frac{(y-3)^2}{9}-\frac{(x+1)^2}{16}=1$ reverses the transverse axis: it would open vertically, with vertices at $y=3\pm3$, and its asymptote slopes would be $\pm\frac{3}{4}$. The choice $\frac{(x+1)^2}{16}-\frac{(y-3)^2}{9}=1$ uses $a=4$ and $b=3$, giving vertices $4$ units from center and slopes $\pm\frac{3}{4}$. The choice $\frac{(x-1)^2}{9}-\frac{(y+3)^2}{16}=1$ has center $(1,-3)$, not $(-1,3)$, because the signs inside the parentheses are reversed. Strategy: find the center, check whether the vertices move horizontally or vertically, then use the asymptote slope $\frac{b}{a}$ to find $b$. This three-step check prevents the most common sign and axis mistakes.

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