Question

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

Answer 1

Answer:

16x^2 - 49y^2 = 784

Step-by-step 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

Answer:

$\frac{x^{2} }{49}-\frac{y^{2} }{16}=1$ is the equation of hyperbola.

Step-by-step 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:

$\frac{x^{2} }{a^{2} }-\frac{y^{2} }{b^{2} }=1$

Substituting a = 7 and b = 4 we get:

$\frac{x^{2} }{49}-\frac{y^{2} }{16}=1$

$16x^{2} -49y^{2}=784$