Question

16: Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the Hyperbola y^2-16x^2=1.

Answer 1

Answer:

Foci $(0,\pm\frac{\sqrt{17}}{4})$

Vertices:$(0,\pm 1)$

Eccentricity, e=$\frac{\sqrt{17}}{4}$

Length of latus rectum=$\frac{1}{8}$

Step-by-step explanation:

We are given that

$y^2-16x^2=1$

$y^2-\frac{x^2}{(\frac{1}{4})^2}=1$

The equation of hyperbola is along y-axis because y is positive

Compare the equation with

$y^2/a^2-x^2/b^2=1$ (Along y-axis)

We get

a=1, b=1/4

$a^2+b^2=c^2$

$1+\frac{1}{16}=c^2$

$\frac{16+1}{16}=c^2$

$c^2=\frac{17}{16}$

$c=\pm \frac{\sqrt{17}}{4}$

Therefore,

The coordinates of foci=$(0,\pm c)=(0,\pm\frac{\sqrt{17}}{4})$

The coordinated of vertices=$(0,\pm a)=(0,\pm 1)$

Eccentricity, e=c/a

$e=\frac{\frac{\sqrt{17}}{4}}{1}=\frac{\sqrt{17}}{4}$

Length of latus rectum=$\frac{2b^2}{a}$

Length of latus rectum=$2\times \frac{1}{16}=\frac{1}{8}$