The equation of the hyperbola with foci (5, 0), (-5, 0) and vertices (4, 0), (-4, 0) is:

Solution:

Given, foci (5,0) (-5,0)

Vertices (4,0) (-4,0)

We have to write the equation of the hyperbola.

There are two types of hyperbolas

1) where a line drawn through its vertices and foci is horizontal

2) where a line drawn through its vertices and foci is vertical

The hyperbola is a type where a line drawn through its vertices and foci is horizontal by observing that x coordinate changes when we move from a focus point to a vertex.

The general equation of this types of hyperbola is

\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}= 1\)

Observing that y coordinate of foci and vertices is 0, this implies that k = 0

Now, the general equation becomes,

\(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-0)^{2}}{b^{2}}= 1\)

The value of h is the sum of x coordinate of the vertices divided by two

h = -4+4/2 = 0

\(\frac{(x-0)^{2}}{a^{2}}-\frac{(y-0)^{2}}{b^{2}}= 1\)

When x = ±4, y = 0. This helps us to find the value of a

\(\frac{(4-0)^{2}}{a^{2}}-\frac{(0-0)^{2}}{b^{2}}= 1\)

\(\frac{(4-0)^{2}}{a^{2}}-0= 1\)

\(\frac{(4-0)^{2}}{a^{2}}= 1\)

\((4-0)^{2}=a^{2}\)

Taking square root,

a = 4

We know, c is the distance between the center point and focus

So, c = 5

\(c^{2}=a^{2}+b^{2}\)

\((5)^{2}=(4)^{2}+b^{2}\)

25 = 16 + b²

b² = 25 - 16

b² = 9

Taking square root,

b = 3

Now, the equation becomes

\(\frac{(x-0)^{2}}{4^{2}}-\frac{(y-0)^{2}}{3^{2}}=1\)

\(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

Therefore, the equation of the hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)

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The equation of the hyperbola with foci (5, 0), (-5, 0) and vertices (4, 0), (-4, 0) is:

Summary:

The equation of the hyperbola with foci (5, 0), (-5, 0) and vertices (4, 0), (-4, 0) is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\)