Answer:
Foci [tex](0,\pm\frac{\sqrt{17}}{4})[/tex]
Vertices:[tex](0,\pm 1)[/tex]
Eccentricity, e=[tex]\frac{\sqrt{17}}{4}[/tex]
Length of latus rectum=[tex]\frac{1}{8}[/tex]
Step-by-step explanation:
We are given that
[tex]y^2-16x^2=1[/tex]
[tex]y^2-\frac{x^2}{(\frac{1}{4})^2}=1[/tex]
The equation of hyperbola is along y-axis because y is positive
Compare the equation with
[tex]y^2/a^2-x^2/b^2=1[/tex] (Along y-axis)
We get
a=1, b=1/4
[tex]a^2+b^2=c^2[/tex]
[tex]1+\frac{1}{16}=c^2[/tex]
[tex]\frac{16+1}{16}=c^2[/tex]
[tex]c^2=\frac{17}{16}[/tex]
[tex]c=\pm \frac{\sqrt{17}}{4}[/tex]
Therefore,
The coordinates of foci=[tex](0,\pm c)=(0,\pm\frac{\sqrt{17}}{4})[/tex]
The coordinated of vertices=[tex](0,\pm a)=(0,\pm 1)[/tex]
Eccentricity, e=c/a
[tex]e=\frac{\frac{\sqrt{17}}{4}}{1}=\frac{\sqrt{17}}{4}[/tex]
Length of latus rectum=[tex]\frac{2b^2}{a}[/tex]
Length of latus rectum=[tex]2\times \frac{1}{16}=\frac{1}{8}[/tex]
