Answer :
Answer:
[tex]\textsf{13.}\quad(x+1)^2-\dfrac{y^2}{3}=1[/tex]
[tex]\textsf{14.}\quad\dfrac{x^2}{16}-\dfrac{y^2}{65}=1[/tex]
[tex]\textsf{15.}\quad y^2-\dfrac{x^2}{9}=1[/tex]
[tex]\textsf{16.}\quad\dfrac{x^2}{36}-\dfrac{y^2}{64}=1[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Standard equation of a vertical hyperbola}\\\\$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center.\\ \phantom{ww}$\bullet$ $(h,k\pm a)$ are the vertices.\\\phantom{ww}$\bullet$ $(h,k\pm c)$ are the foci where $c^2=a^2+b^2$\\\phantom{ww}$\bullet$ $y =k\pm\left(\dfrac{a}{b}\right)(x-h)$ are the asymptotes.\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Standard equation of a horizontal hyperbola}\\\\$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$\\\\where:\\\phantom{ww}$\bullet$ $(h,k)$ is the center.\\ \phantom{ww}$\bullet$ $(h\pm a,k)$ are the vertices.\\\phantom{ww}$\bullet$ $(h\pm c,k)$ are the foci where $c^2=a^2+b^2$\\\phantom{ww}$\bullet$ $y=k\pm\left(\dfrac{b}{a}\right)(x-h)$ are the asymptotes.\\\end{minipage}}[/tex]
Question 13
Given:
- foci: (-3, 0) and (1, 0)
- vertices: (-2, 0) and (0, 0)
As the y-values of the foci and vertices are the same, the hyperbola is horizontal (opening left and right).
The center (h, k) is the midpoint of the vertices.
Therefore, the center is (-1, 0) and so:
- h = -1
- k = 0
Use the formula for the vertices (h±a, k) to determine the value of a:
[tex]\begin{aligned}\implies h\pm a&=-2\\-1\pm a&=-2\\\pm a&=-1\end{aligned}[/tex] [tex]\begin{aligned}\implies h\pm a&=0\\-1\pm a&=0\\\pm a&=1\end{aligned}[/tex]
Therefore:
- a² = 1
Use the formula for the foci (h±c, k) to determine the value of c:
[tex]\begin{aligned}\implies h\pm c&=-3\\-1\pm c&=-3\\\pm c&=-2\end{aligned}[/tex] [tex]\begin{aligned}\implies h\pm c &=1\\-1\pm c&=1\\\pm c &=2\end{aligned}[/tex]
Therefore:
- c = 2
To find b² use c² = a² + b² and the found values of a and c:
[tex]\begin{aligned}\implies c^2&=a^2+b^2\\2^2&=1^2+b^2\\4&=1+b^2\\b^2&=3\end{aligned}[/tex]
Substitute the found values of h, k, a² and b² into the formula to create an equation of the hyperbola:
[tex]\implies\dfrac{(x+1)^2}{1}-\dfrac{(y-0)^2}{3}=1[/tex]
[tex]\implies(x+1)^2-\dfrac{y^2}{3}=1[/tex]
Question 14
Given:
- foci: (-9, 0) and (9, 0)
- vertices: (-4, 0) and (4, 0)
As the y-values of the foci and vertices are the same, the hyperbola is horizontal.
The center (h, k) is the midpoint of the vertices.
Therefore, the center is (0, 0) and so:
- h = 0
- k = 0
Use the formula for the vertices (h±a, k) to determine the value of a:
[tex]\begin{aligned}\implies h\pm a&=\pm4\\0\pm a&=\pm4\\\pm a&=\pm4\end{aligned}[/tex]
Therefore:
- a² = 16
Use the formula for the foci (h±c, k) to determine the value of c:
[tex]\begin{aligned}\implies h \pm c&=\pm9\\0\pm c&=\pm9\\\pm c&=\pm9\end{aligned}[/tex]
Therefore:
- c = 9
To find b² use c² = a² + b² and the found values of a and c:
[tex]\begin{aligned}\implies c^2&=a^2+b^2\\9^2&=4^2+b^2\\81&=16+b^2\\b^2&=65\end{aligned}[/tex]
Substitute the found values of h, k, a² and b² into the formula to create an equation of the hyperbola:
[tex]\implies\dfrac{(x-0)^2}{16}-\dfrac{(y-0)^2}{65}=1[/tex]
[tex]\implies\dfrac{x^2}{16}-\dfrac{y^2}{65}=1[/tex]
Question 15
Given:
- vertices: (0, -1) and (0, 1)
- [tex]\textsf{asymptotes}:\;\;y=\dfrac{1}{3}x\;\;\textsf{and}\;\;y=-\dfrac{1}{3}x[/tex]
As the x-values of the vertices are the same, the hyperbola is vertical (opening up and down).
The center (h, k) is the midpoint of the vertices.
Therefore, the center is (0, 0) and so:
- h = 0
- k = 0
Use the formula for the vertices (h, k±a) to determine the value of a:
[tex]\begin{aligned}\implies k \pm a&=\pm1\\0\pm a&=\pm1\\\pm a&=\pm1\end{aligned}[/tex]
Therefore:
- a² = 1
Use the formula for the asymptotes to determine the value of b:
[tex]\begin{aligned}\implies k\pm\left(\dfrac{a}{b}\right)(x-h)&=\pm\dfrac{1}{3}x\\\\0\pm \left(\dfrac{1}{b}\right)(x-0)&=\pm\dfrac{1}{3}x\\\\\pm\dfrac{1}{b}x&=\pm\dfrac{1}{3}x\\\\\pm b&=\pm3\end{aligned}[/tex]
Therefore:
- b² = 9
Substitute the found values of h, k, a² and b² into the formula to create an equation of the hyperbola:
[tex]\implies\dfrac{(y-0)^2}{1}-\dfrac{(x-0)^2}{9}=1[/tex]
[tex]\implies y^2-\dfrac{x^2}{9}=1[/tex]
Question 16
As the y-values of the vertices are the same, the hyperbola is horizontal.
The center (h, k) is the midpoint of the vertices.
Therefore, the center is (0, 0) and so:
- h = 0
- k = 0
Use the formula for the vertices (h±a, k) to determine the value of a:
[tex]\begin{aligned}\implies h \pm a&=\pm6\\0\pm a&=\pm6\\\pm a&=\pm6\end{aligned}[/tex]
Therefore:
- a² = 36
Use the formula for the asymptotes to determine the value of b:
[tex]\begin{aligned}\implies k \pm\left(\dfrac{b}{a}\right)(x-h)&=\pm\dfrac{4}{3}x\\\\0\pm \left(\dfrac{b}{6}\right)(x-0)&=\pm\dfrac{4}{3}x\\\\\pm \dfrac{b}{6}x&=\pm\dfrac{4}{3}x\\\\ \pm b&=\pm8\end{aligned}[/tex]
Therefore:
- b² = 64
Substitute the found values of h, k, a² and b² into the formula to create an equation of the hyperbola:
[tex]\implies \dfrac{x^2}{36}-\dfrac{y^2}{64}=1[/tex]
