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

Foci
`(0,\pm(√(17))/(4))`

Vertices:
`(0,\pm 1)`

Eccentricity, e=
`(√(17))/(4)`

Length of latus rectum=
`(1)/(8)`

Explanation:

We are given that

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

`y^2-(x^2)/(((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+(1)/(16)=c^2`

`(16+1)/(16)=c^2`

`c^2=(17)/(16)`

`c=\pm (√(17))/(4)`

Therefore,

The coordinates of foci=
`(0,\pm c)=(0,\pm(√(17))/(4))`

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

Eccentricity, e=c/a

`e=((√(17))/(4))/(1)=(√(17))/(4)`

Length of latus rectum=
`(2b^2)/(a)`

Length of latus rectum=
`2* (1)/(16)=(1)/(8)`