Question

Write the equation of a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2(sqrt5)

Answer 1

`(x^2)/(16)-(b^2)/(4)=1`

Explanation:

A hyperbola is the locus of a point such that its distance from a point to two points (known as foci) is a positive constant.

The standard equation of a hyperbola centered at the origin with transverse on the x axis is given as:

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

The coordinates of the foci is at (±c, 0), where c² = a² + b²

Given that a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2√5. Since the x intercept is ±4, this means that at y = 0, x = 4. Substituting in the standard equation:

`(x^2)/(a^2)-(y^2)/(b^2)=1\\(4^2)/(a^2)-(0)/(b^2) =1\\(4^2)/(a^2)=1\\ a^2=16\\a=√(16)=4\\ a=4`

The foci c is at +/-2√5, using c² = a² + b²:

`c^2=a^2+b^2\\(2√(5) )^2=4^2+b^2\\20 = 16 + b^2\\b^2=20-16\\b^2=4\\b=√(4)=2\\ b=2`

Substituting the value of a and b to get the equation of the hyperbola:

`(x^2)/(a^2)-(y^2)/(b^2)=1\\\\(x^2)/(16)-(b^2)/(4)=1`