A hyperbola is the set of all points in a plane such that the difference of whose distances from two distinct fixed points called foci is a positive constant. In this problem, we have the following equation:
[tex]49x^2-100y^2=4900[/tex]
What if we divide the whole equation by [tex]49 \times 100=4900[/tex]? Well the result is:
[tex]\frac{1}{4900}(49x^2-100y^2=4900) \\ \\ \frac{49x^2}{4900}-\frac{100y^2}{4900}=\frac{4900}{4900} \\ \\ \frac{x^2}{100}-\frac{y^2}{49}=1 \\ \\ \boxed{\frac{x^2}{10^2}-\frac{y^2}{7^2}=1}[/tex]
The standard form of the equation of the hyperbola given that the vertex lies on the origin is:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
SO THE VERTICES ARE:
[tex]\boxed{(a,0)=(10,0) \ and \ (-a,0)=(-10,0)}[/tex]
Calculating the foci:
[tex]We \ know \ that \ foci \ are: \\ \\ (-c,0) \ and \ (c.0) \\ \\ Also: \\ \\ c=\sqrt{a^2+b^2} \\ \\ c=\sqrt{100+49} \therefore c=\sqrt{149}[/tex]
SO THE FOCI ARE:
[tex]\boxed{(-\sqrt{149},0) \ and \ (\sqrt{149},0)}[/tex]
