Question

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

Answer 1

Answer:

$\frac{x^2}{16}-\frac{b^2}{4}=1$

Step-by-step 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:

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

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:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\\frac{4^2}{a^2}-\frac{0}{b^2} =1\\\frac{4^2}{a^2}=1\\ a^2=16\\a=\sqrt{16}=4\\ a=4$

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

$c^2=a^2+b^2\\(2\sqrt{5} )^2=4^2+b^2\\20 = 16 + b^2\\b^2=20-16\\b^2=4\\b=\sqrt{4}=2\\ b=2$

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

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\\\\frac{x^2}{16}-\frac{b^2}{4}=1$