Find an equation for the ellipse with foci at (0, -2) and (0, 2); length of the major axis is 8?

Question

Find an equation for the ellipse with vertices at (-3, 4) and (15, 4); focus at (13, 4)

Find an equation for the hyperbola with vertices at (0, -6) and (0, 6); asymptote the line y = 3/5x

Find an equation for the hyperbola with center at (8, 2); focus at (5, 2); vertex at (7, 2)

Answer 1

2.

The vertices are (-3,4) and(15,4).

Focus is at (13,4).

midpoint =[(-3+15)/2,(4+4)/2] =(6,4).

Hence (h,k) = (6,4).

The distance between centre and foci is 

(13,4),(6,4)

Hence the value of c = 7.

distance from centre to one of the vertex is 

(6,4),(-3,4)

So,the value of a=9.

From,

 

The equation of ellipse is 

 

So, the equation of ellipse is 

Answer 2

1.

The foci is at (0,-2)and (0,2) that is (0,± 2).

Length of major axis (2a) =8

=>a=4.

So,the value of c=2

from 

The equation for the ellipse is 

Hence the equation of the ellipse is  .

Answer 3

3.

The vertices are (0,-6) and (0,6)

The asymptote is y=(3/5)x

Here the vertices are in the form of (0, ±a)   that is (0,±6).

So, it is vertical hyperbola and the equation for vertical hyperbola is .

Here a = 6 and from the asymptote line equation m = 3/5.

m= a / b =6 / b = 3/5.

Hence, b= 10.

So,the equation for the hyperbola is .

Hence the equation of hyperbola is .

Answer 4

4.

The Centre is at(8,2)  and focus at (5,2) and vertex at (7,2)

The Centre, focus, vertex all lie on  horizontal line y=2.

Distance from Centre to vertex :

(8,2) (7,2)

Hence the vertex is 1 unit from the Centre that is a=1.

Distance from Centre to focus :

(8,2)(5,2)

Hence the focus is 3 units from the Centre that is c=3.

From 

Now ,the equation of the hyperbola is .

=>.

Hence the equation of the hyperbola is .