Question

HELP ASAP: Find a hyperbola's equation with vertices at (-7, 0) + (7, 0) and co-vertices (0, -4) + (0, 4).

Answer 1

16x^2 - 49y^2 = 784

Explanation:

as we know that this is the standard equation of hyperbola bcz 1 coordinate of vertices is 0.

since in vertices x coordinate is present and y coordinate is 0 so Tranverse axis of hyperbola is along x axis .

vertices in general = (a,0) and (-a,0)

vertices given = ( 7,0) and (-7 , 0)

so a=7

similarly

co vertices in general = ( 0 , b) and ( 0 , -b)

co vertices given = ( 0, 4) and (0 , -4)

so,

b= 4

now equation for x axis is:

(x2/a2) - ( y2/b2) = 1

by putting values we observe the ans is

16x^2 - 49y^2 = 784

Answer 2

`(x^(2) )/(49)-(y^(2) )/(16)=1` is the equation of hyperbola.

Explanation:

Given:

Vertices of the hyperbola: (-7,0) and (7,0)

Co-vertices of the hyperbola: (0,-4) and (0,4)

Mid-point of the vertices = center of hyperbola = (0,0)

Focii lie on the same line as vertices and hence they lie on x-axis.

Here x-axis is the tranverse axis and y-axis is the conjugate axis.

length of semi-transverse axis = a = 7

length of semi-conjugate axis = b = 4

Equation of hyperbola is of the form:

`(x^(2) )/(a^(2) )-(y^(2) )/(b^(2) )=1`

Substituting a = 7 and b = 4 we get:

`(x^(2) )/(49)-(y^(2) )/(16)=1`

`16x^(2) -49y^(2)=784`