Question

0

Answer 1

recall that the equation of a hyperbola looks like \(\bf \cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\\
\cfrac{(y-k)^2}{a^2}-\cfrac{(x-h)^2}{b^2}=1\)

Answer 2

so \(\bf 4x^2 - y^2 = 100 \implies \cfrac{4x^2}{100}-\cfrac{y^2}{100}=1 \implies \cfrac{x^2}{25}-\cfrac{y^2}{100}=1\\ \quad \\
\color{blue}{25 = 5^2\qquad \qquad 100 = 10^2}\\ \quad \\
\cfrac{x^2}{5^2}-\cfrac{y^2}{10^2}=1 \implies \cfrac{(x-0)^2}{5^2}-\cfrac{(y-0)^2}{10^2}=1 \)

Answer 3

this hyperbola, notice the fraction with the POSITIVE sign, the "fraction with the x", thus the hyperbola opens towards the x-axis, that is, to the left and right

you can get the asymptotes for a "horizontally traverse axis" hyperbola at \(\bf y = k\pm \cfrac{b}{a}(x-h)\)

Answer 4

so that's 0+- 25/100(x-0)??

Answer 5

\(\bf \bf y = k\pm \cfrac{b}{a}(x-h)\) <---- notice "a" and "b", not squared

Answer 6

your hyperbola will be like \(\bf \cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\)

center is at (h, k)

Answer 7

isnt that just 0,0 then in what you typed in?

Answer 8

yea

Answer 9

how do i find the asymptotes ?

Answer 10

so your asymptotes will be at \(\bf \cfrac{(x-0)^2}{5^2}-\cfrac{(y-0)^2}{10^2}=1\\ \quad \\ \quad \\
y = k\pm \cfrac{b}{a}(x-h) \implies y = 0\pm \cfrac{10}{5}(x-0)\)

Answer 11

can you explain how to plug that in? if i only have h and k?

Answer 12

\(\bf \cfrac{(x-0)^2}{\color{red}{5}^2}-\cfrac{(y-0)^2}{\color{red}{10}^2}=1 \implies \cfrac{(x-h)^2}{\color{red}{a}^2}-\cfrac{(y-k)^2}{\color{red}{b}^2}=1\\ \quad \\ \quad \\
y = k\pm \cfrac{\color{red}{b}}{\color{red}{a}}(x-h) \implies y = 0\pm \cfrac{10}{5}(x-0)\)

Answer 13

asymptotes 5/10 so y= 1/2x and y= -1/2x? center (0,0)?

Answer 14

5/10?

Answer 15

notice, the "b" is in the numerator

Answer 16

10/5* = y=2x and y=-2x?

Answer 17

yeap

Answer 18

ty

Answer 19

yw