Question

((y+5)^2 )/16 - (x^2)/24 = 1

Find the center, foci, vertices, and asymptotes of the hyperbola

There should be 2 vertices, 2 foci, and 2 asymptote equations.

Answer 1

`\textsf{Center:}\quad (0, -5)`

`\textsf{Foci:}\quad \left(0,-5-2√(10)\right)\;\textsf{and}\;\left(0,-5+2√(10)\right)`

`\textsf{Vertices:}\quad (0, -9)\; \textsf{and}\;(0,-1)`

`\textsf{Asymptotes:} \quad y=(√(6))/(3)x-5\;,\quad y= -(√(6))/(3)x-5`

Explanation:

Given equation of the hyperbola:

`((y+5)^2)/(16)-(x^2)/(24)=1`

Since the y-term is positive, the hyperbola is vertical and opens up and down.

The standard equation of a vertical hyperbola is:

`\boxed{\begin{minipage}{7.4 cm}\underline{Standard equation of a vertical hyperbola}\\\\$((y-k)^2)/(a^2)-((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}$\bullet$ $(h\pm b, k)$ are the co-vertices. \\\phantom{ww}$\bullet$ $(h, k\pm c)$ are the foci where $c^2=a^2+b^2$\\\phantom{ww}$\bullet$ $y =k \pm \left((a)/(b)\right)(x-h)$ are the asymptotes.\\\end{minipage}}`

Comparing the given equation with the standard equation, we have:

• 
  `h = 0`
• 
  `k = -5`
• 
  `a^2 = 16 \implies a=√(16)=4`
• 
  `b^2 = 24 \implies b=√(24)=2√(6)`

Therefore, the center (h, k) is (0, -5).

To find the foci, we first need to find the value of c by substituting the values of a² and b² into the equation c² = a² + b²:

`\begin{aligned}c^2&=a^2+b^2\\c&=√(a^2+b^2)\\c&=√(16+24)\\c&=√(40)\\c&=√(2^2 \cdot 10)\\c&=√(2^2) √(10)\\c&=2√(10)\end{aligned}`

Now, to determine the foci, substitute the values of h, k and c into the foci formula:

`\begin{aligned}\sf Foci&=(h,k\pm c)\\&=\left(0,-5\pm 2√(10)\right)\end{aligned}`

To find the coordinates of the vertices, substitute the values of h, k and a into the vertices formula:

`\begin{aligned}\textsf{Vertices}&=(h, k \pm a)\\&=(0, -5 \pm 4)\\&=(0, -9)\; \textsf{and}\;(0,-1)\end{aligned}`

To find the equations of the asymptotes, substitute the values of h, k, a and b into the asymptote formula:

`\begin{aligned}y &=k \pm \left((a)/(b)\right)(x-h)\\\\y &=-5 \pm \left((4)/(2√(6))\right)(x-0)\\\\y &=-5 \pm \left((4)/(2√(6))\right)x\\\\y &=-5 \pm \left((2)/(√(6))\right)x\\\\y &=-5 \pm \left((2√(6))/(√(6)√(6))\right)x\\\\y &=-5 \pm \left((2√(6))/(6)\right)x\\\\y &=-5 \pm (√(6))/(3)x\end{aligned}`

Therefore, the equations of the asymptotes are:

`y=(√(6))/(3)x-5`

`y= -(√(6))/(3)x-5`

Answer 2

centre = (0,-5)

vertices are:

`\sf \textsf{Vertex 1: } (0,-1)`

`\sf \textsf{Vertex 2: } (0, - 9)`

Foci:

`\sf \textsf{Focus 1:} (0, -5 + 2√(10) )`

`\sf \textsf{Focus 2:} (0, -5 - 2√(10) )`

Equations of the asymptotes:

`\sf y =(√(6))/(3)x-5`

`\sf y = - (√(6))/(3)x-5`

Explanation:

`\sf ((y+5)^2)/(16) - (x^2)/(24)=1`

The given equation, represents a hyperbola. To identify its vertices, foci, and equations of asymptotes, we first need to rewrite the equation in standard form, which is of the form:

For a horizontal hyperbola

`\sf ((x - h)^2 )/( a^2) - ((y - k)^2 )/(b^2) = 1`

For a vertical hyperbola

`\sf ((y - k)^2 )/( b^2) - ((x - h)^2 )/(a^2) = 1`

Where:

(h, k) is the center of the hyperbola.

'a' is the distance from the center to the vertices along the major axis.

'b' is the distance from the center to the co-vertices along the minor axis.

In this case, the equation is in standard form, so

Comparing the coefficients with standard form for a vertical hyperbola. we have:

• h = 0
• k = -5
• a² = 24
• b² = 16

Now that we have the values of 'h', 'k', 'a', and 'b', we can find the vertices, foci, and equations of the asymptotes.

Centre:

Centre are located in central point. Centre point is given by (h,k).

Therefore, centre = (h,k) = (0,-5)

Vertices:

The vertices are located along the major axis. For a vertical hyperbola, the vertices are given by (h , k ± b), which in this case are:

`\sf ( 0,-5 \pm √(16) )`

`\sf ( 0, -5 \pm 4 )`

So the vertices are:

`\sf \textsf{Vertex 1: } (0,-1)`

`\sf \textsf{Vertex 2: } (0, - 9)`

Foci:

The distance from the center to the foci is given by c, where

`\sf c^2 = a^2 + b^2`

In this case,

`\sf c^2 = 24 + 16`

`\sf c^2 = 40`

`\sf c = √(40)`

`\sf c = 2√(10)`

The foci are located along the major axis, so their coordinates are (h , k ± c ):

`\sf (0, -5 \pm 2√(10) )`

So,

`\sf \textsf{Focus 1:} (0, -5 + 2√(10) )`

`\sf \textsf{Focus 2:} (0, -5 - 2√(10) )`

Equations of Asymptotes:

The equations of the asymptotes for a vertical hyperbola are given by:

`y = \pm (b)/(a)(x - h) + k`

Substituting the value, we get

`\sf y = \pm (4)/(√(24))(x - 0) -5`

`\sf y = \pm (4)/(2√(6))x -5`

`\sf y = \pm (4)/(2√(6))x * (√(6))/(√(6))-5`

`\sf y = \pm (2 √(6))/(6)x-5`

`\sf y = \pm (√(6))/(3)x-5`

So, the equations of the asymptotes are:

`\sf y =(√(6))/(3)x-5`

`\sf y = - (√(6))/(3)x-5`