nagasenaz

2020-12-03

Instructions: Graph the conic section and make sure to label the coordinates in the graph. Include all the calculations needed to complete the graph. Give the standard form (SF) and the general form (GF) of the conic sections.HYPERBOLA:1) The vertices are at (-2, 0) and (2, 0). The conjugate axis' length is 6.

Derrick

Given: The vertices are at (-2, 0) and (2, 0). The conjugate axis's length is 6. The two vertices have the same y coordinate, therefore, the hyperbola is the horizontal transverse axis type. The standard equation for the horizontal transverse axis type is $\frac{\left(x-h{\right)}^{2}}{{a}^{2}}-\frac{\left(y-k{\right)}^{2}}{{b}^{2}}=1$ We know that the general form for the vertices of this type is: comparing with vertices $k=0$
$h+a=2$
$h-a=-2$ Solve $h=0,a=2$ The length of the conjugate axis is equal to 2b $2b=6$
$b=3$ Standard form $\frac{\left(x-0{\right)}^{2}}{{2}^{2}}+\frac{\left(y-{0}^{2}\right)}{{3}^{2}}=1$
$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$ General form $9{x}^{2}+4{y}^{2}=36$
$9{x}^{2}+4{y}^{2}-36=0$

Do you have a similar question?