Answer:
a)
The vertices are [tex]\left(3,\:0\right),\:\left(-3,\:0\right)[/tex].
The foci are [tex]\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)[/tex].
The asymptotes are [tex]y=2x,\:y=-2x[/tex].
b) The length of the transverse axis is 6.
c) See below.
Step-by-step explanation:
[tex]\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1[/tex] is the standard equation for a right-left facing hyperbola with center [tex]\left(h,\:k\right)[/tex].
a)
The vertices[tex]\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)[/tex] are the two bending points of the hyperbola with center [tex]\:\left(h,\:k\right)[/tex] and semi-axis a, b.
Therefore,
[tex]\frac{x^2}{9}-\frac{y^2}{36}=1[/tex], is a right-left Hyperbola with [tex]\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6[/tex] and vertices [tex]\left(3,\:0\right),\:\left(-3,\:0\right)[/tex].
For a right-left facing hyperbola, the Foci (focus points) are defined as [tex]\left(h+c,\:k\right),\:\left(h-c,\:k\right)[/tex] where [tex]c=\sqrt{a^2+b^2}[/tex] is the distance from the center [tex]\left(h,\:k\right)[/tex] to a focus.
Therefore,
[tex]\frac{x^2}{9}-\frac{y^2}{36}=1[/tex], is a right-left Hyperbola with [tex]\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6[/tex] [tex]c=\sqrt{3^2+6^2}= 3\sqrt{5}[/tex] and foci [tex]\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)[/tex]
The asymptotes are the lines the hyperbola tends to at [tex]\pm \infty[/tex]. For right-left hyperbola the asymptotes are: [tex]y=\pm \frac{b}{a}\left(x-h\right)+k[/tex]
Therefore,
[tex]\frac{x^2}{9}-\frac{y^2}{36}=1[/tex], is a right-left Hyperbola with [tex]\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6[/tex] and asymptotes
[tex]y=\frac{6}{3}\left(x-0\right)+0,\:\quad \:y=-\frac{6}{3}\left(x-0\right)+0\\y=2x,\:\quad \:y=-2x[/tex]
b) The length of the transverse axis is given by [tex]2a[/tex]. Therefore, the lenght is 6.
c) See below.
