Answer
See graph
Foci: $(-\sqrt{53},0),(\sqrt{53},0)$
Work Step by Step
We are given the hyperbola:
$4x^2-49y^2=196$
Bring the equation to the standard form:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
$\dfrac{4x^2}{196}-\dfrac{49y^2}{196}=1$
$\dfrac{x^2}{49}-\dfrac{y^2}{4}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=49\Rightarrow a=\sqrt {49}=7$
$b^2=4\Rightarrow b=\sqrt {4}=2$
$c^2=a^2+b^2$
$c^2=49+4$
$c^2=53$
$c=\sqrt{53}$
The centre of the hyperbola is:
$(h,k)=(0,0)$
Determine the coordinates of the vertices:
$(h-a,k)=(0-7,0)=(-7,0)$
$(h+a,k)=(0+7,0)=(7,0)$
Determine the coordinates of the foci:
$(h-c,k)=(0-\sqrt{53},0)=(-\sqrt{53},0)$
$(h+c,k)=(0+\sqrt{53},0)=(\sqrt{53},0)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y-0=\pm\dfrac{2}{7}(x-0)$
$y=\pm\dfrac{2}{7}x$
Graph the hyperbola:
