Question

Identify the vertices, foci, and equations of the asymptotes of the hyperbola.

(y^2)/24 - (x^2)/18 = 1

Answer 1

Answer:

$\textsf{Vertices:}\quad (0,- 2\sqrt{6})\;\;\textsf{and}\;\;(0,2\sqrt{6})$

$\textsf{Co-vertices:}\quad (-3\sqrt{2},0)\;\;\textsf{and}\;\;(3\sqrt{2},0)$

$\textsf{Foci:}\quad (0,-\sqrt{42})\;\;\textsf{and}\;\;(0,\sqrt{42})$

$\textsf{Asymptotes:}\quad y =\dfrac{2\sqrt{3}}{3}x\;\;\textsf{and}\;\;y =-\dfrac{2\sqrt{3}}{3}x$

Step-by-step explanation:

Given equation of a hyperbola:

$\dfrac{y^2}{24}-\dfrac{x^2}{18}=1$

As the y²-term of the given equation is positive, the transverse axis is vertical, and so the hyperbola is vertical (opening up and down).

The general equation of a vertical hyperbola is:

$\boxed{\begin{array}{l}\underline{\textsf{Vertical hyperbola}}\\\\\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1\\\\\textsf{where:}\\\phantom{ww}\;\bullet\;\textsf{ (h,k) is the center.}\\\phantom{ww}\;\bullet\;\textsf{ (h, k\pm a) are the vertices.}\\\phantom{ww}\;\bullet\;\textsf{ (h\pm b, k) are the co-vertices.}\\\phantom{ww}\;\bullet\;\textsf{ (h, k\pm c) are the foci where c^2=a^2+b^2 }\\\phantom{ww}\;\bullet\;\textsf{ y =k\pm\left(\dfrac{a}{b}\right)(x-h) are the asymptotes.}\end{array}}$

In this case:

$h = 0$

$k = 0$

$a^2 = 24\implies a=\sqrt{24}=2\sqrt{6}$

$b^2 = 18\implies b=\sqrt{18}=3\sqrt{2}$

Substitute the values of h, k, a and b into the formulas for the vertices and co-vertices:

$\textsf{Vertices:}\quad (h, k\pm a)&=(0,\pm 2\sqrt{6})$

$\textsf{Co-vertices:}\quad (h\pm b,k)&=(\pm 3\sqrt{2},0)$

Find the value of c:

$c^2=a^2+b^2$

$c^2=24+18$

$c^2=42$

$c=\sqrt{42}$

Substitute the values of h, k, and c into the formula for the foci:

$\textsf{Foci:}\quad (h, k\pm c)&=(0,\pm \sqrt{42})$

Substitute the values of h, k, a and b into the formulas for the asymptotes:

$y =0\pm\left(\dfrac{2\sqrt{6}}{3\sqrt{2}}\right)(x-0)$

$y =\pm\left(\dfrac{2\sqrt{6}\cdot \sqrt{2}}{3\sqrt{2}\cdot \sqrt{2}}\right)x$

$y =\pm\left(\dfrac{2\sqrt{12}}{3\cdot 2}\right)x$

$y =\pm\left(\dfrac{2\sqrt{2^2\cdot 3}}{3\cdot 2}\right)x$

$y =\pm\left(\dfrac{2\sqrt{2^2}\sqrt{3}}{3\cdot 2}\right)x$

$y =\pm\left(\dfrac{2\cdot 2\sqrt{3}}{3\cdot 2}\right)x$

$y =\pm\dfrac{2\sqrt{3}}{3}x$

Answer 2

Answer:

Vertices: $(0, 2\sqrt{6}) \textsf{ and} (0, -2\sqrt{6})$

Foci: $(0, \sqrt{42})\textsf{ and } (0, -\sqrt{42})$

Equation of asymptotes: $y = \dfrac{2\sqrt{3}{3}}{3}x \textsf{ and } y = - \dfrac{2\sqrt{3}{3}}{3}x$

Step-by-step explanation:

The equation $\dfrac{y^2}{24} - \dfrac{x^2}{18} = 1$ is in the standard form of a hyperbola:

$\Large\boxed{\boxed{ \dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1}}$

where

• $a$ and $b$ are the lengths of the transverse and conjugate axes, respectively.

For the equation $\dfrac{y^2}{24} - \dfrac{x^2}{18} = 1$:

• $a^2 = 18$ (since $a^2$ is the denominator of $x^2$),
• $b^2 = 24$ (since $b^2$ is the denominator of $y^2$).

This hyperbola has a vertical transverse axis because the term involving $y^2$ is positive.

Vertices:

The vertices are the points where the hyperbola intersects its transverse axis. Since this hyperbola has a vertical transverse axis, the vertices are along the $y$-axis.

The vertices are given by $(0, \pm b)$ .so in our case, the vertices are $(0, \pm \sqrt{24}) = (0, \pm 2\sqrt{6} )$.

Two vertices are:

$(0, 2\sqrt{6}) \textsf{ and} (0, -2\sqrt{6})$

Foci:

The distance from the center to the foci is given by $c = \sqrt{a^2 + b^2}$.

In our case:

$c = \sqrt{18 + 24}\\\\ = \sqrt{42}$

The foci are located along the transverse axis, so they're $c$ units above and below the center.

Therefore, the foci are at $(0, \sqrt{42})\textsf{ and } (0, -\sqrt{42})$.

Asymptotes:

The asymptotes of a hyperbola have the equation:

$y = \pm \dfrac{b}{a}x$

In our case, the equation becomes:

$\begin{aligned} y &= \pm \dfrac{\sqrt{24}}{\sqrt{18}}x \\\\&= \pm \dfrac{\sqrt{2^2\cdot 6}}{\sqrt{3^2 \cdot 2}}x \\\\&= \pm \dfrac{2\sqrt{6}}{3\sqrt{2}}x \\\\&= \pm \dfrac{2\sqrt{6}}{3\sqrt{2}}x \cdot \dfrac{\sqrt{2}}{\sqrt{2}} \\\\&= \pm \dfrac{2\sqrt{6}\cdot \sqrt{2}}{3 \sqrt{2}\cdot \sqrt{2}} x \\\\&= \pm \dfrac{2\sqrt{6\cdot 2}}{3 \sqrt{2\cdot 2}}x \\\\&= \pm \dfrac{\cancel{2}\sqrt{2^2 \cdot 3}}{3\cdot \cancel{2}} x \\\\&= \pm \dfrac{2\sqrt{3}}{3}x \end{aligned}$

Thus, the equations of the asymptotes are $y = \dfrac{2\sqrt{3}{3}}{3}x \textsf{ and } y = - \dfrac{2\sqrt{3}{3}}{3}x$.