Show each step of your work and fully simplify each expression.
Turn in your answers in class on a physical piece of paper.
Staple multiple sheets together.
Feel free to use Desmos for graphing.
Answer the following:
For each of the following problems, find an equation of the hyperbola with vertex at the origin satisfying the following conditions or state it's impossible to.
Foci $(\pm 5, 0)$, vertices $(\pm 3, 0)$
Foci $(0, \pm \sqrt{10})$, vertices $(\pm 1, 0)$
Asymptotes $y = \pm x$, vertices $(0, \pm 3)$
Length of transverse axis is 6, oriented vertically.
Hint: find the vertices.
Foci $(\pm 5, 0)$, length of transverse axis is 6
This hyperbola:
Consider the hyperbola \[\dfrac{x^2}{4} - \dfrac{y^2}{8} = 1\] If I want to shift it three units right and four units down, what is the new formula?
For the following ellipses, sketch the graph and find the foci.
$\dfrac{x^2}{9} + \dfrac{(y+5)^2}{25} = 1$
$\dfrac{(x + 1)^2}{36} + \dfrac{(y+1)^2}{64} = 1$
$25(x-1)^2 + 9(y-2)^2 = 25 * 9$
For the following parabolas, sketch the graph and find the focus.
$(x-3)^2 = 8(y+1)$
$(y+1)^2 = 16(x - 3)$
$y^2 - 6y - 12x + 33 = 0$
$x^2 + 2x - 20y + 41 = 0$
For the following hyperbolas, sketch the graph and find the foci.