Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±five divided by four times x..

so, based on the provided vertices, the hyperbola will look more or less like the picture below, with those asymptotes, so is a vertical hyperbola.

since the center is at the origina, that makes the traverse axis of 20, namely a = 10, what is "b" then?

[tex]\bf \textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ asymptotes\quad y= k\pm \cfrac{a}{b}(x- h) \end{cases}\\\\ -------------------------------[/tex]

[tex]\bf \pm\cfrac{5}{4}x~~=~~k\pm \cfrac{a}{b}(x- h)\implies \pm\cfrac{5}{4}x~~=~~0\pm \cfrac{a}{b}(x- 0) \\\\\\ \pm\cfrac{5}{4}x~~=~~\pm\cfrac{a}{b}x\impliedby \begin{array}{llll} \textit{let us use the positive one}\\ \textit{and since we know }a=10 \end{array} \\\\\\ \cfrac{5x}{4}=\cfrac{10x}{b}\implies \cfrac{5x}{10x}=\cfrac{4}{b}\implies \cfrac{1}{2}=\cfrac{4}{b}\implies b=8\\\\ -------------------------------\\\\ \cfrac{(y-0)^2}{10^2}-\cfrac{(x-0)^2}{8^2}=1\implies \cfrac{y^2}{100}-\cfrac{x^2}{64}=1[/tex]

erinna erinna · Answer 2 · 2019-08-19T11:14:21+02:00

Answer:

The equation of hyperbola is [tex]\frac{y^2}{10^2}-\frac{x^2}{8^2}=1[/tex].

Step-by-step explanation:

Given information: Vertices at (0, ±10) asymptotes at [tex]y=\pm \frac{5}{4}x[/tex].

Vertices are on the y-axis, so given hyperbola is along the y-axis. The standard form for the hyperbola is

[tex]\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1[/tex]

where, (h,k) is the center of hyperbola. (0,±b) are vertex and [tex]y=\pm \frac{b}{a}x[/tex] are asymptotes.

Vertices at (0, ±10), it means center of the hyperbola is (0,0). So the standard form for the hyperbola is

[tex]\frac{y^2}{b^2}-\frac{x^2}{a^2}=1[/tex] .... (1)

Vertices of hyperbola:

[tex](0,\pm b)=(0, \pm 10)[/tex]

On comparing we get

[tex]b=10[/tex]

The value of b is 10.

Asymptotes of the hyperbola:

[tex]\pm \frac{b}{a}x=\pm \frac{5}{4}x[/tex]

On comparing both sides, we get

[tex]\frac{b}{a}=\frac{5}{4}[/tex]

Substitute b=10.

[tex]\frac{10}{a}=\frac{5}{4}[/tex]

On cross multiplication,

[tex]40=5a[/tex]

Divide both sides by 5.

[tex]8=a[/tex]

The value of a is 8.

Substitute a=8 and b=10 in equation (1) to find the equation of hyperbola.

[tex]\frac{y^2}{10^2}-\frac{x^2}{8^2}=1[/tex]

Therefore the equation of hyperbola is [tex]\frac{y^2}{10^2}-\frac{x^2}{8^2}=1[/tex].