# Diameter of a Hyperbola

I. Introduction to Diameter:

• In geometry, the diameter of a geometric figure is a line segment connecting two points on the curve and passing through the center. For a hyperbola, the concept of a diameter is specific and noteworthy.

II. Diameter of a Hyperbola:

1. Definition:

• The diameter of a hyperbola is a line segment passing through the center and having both endpoints on the hyperbola.
• It is the longest chord of the hyperbola.
2. Length:

• The length of the diameter is twice the length of the semi-major axis.
• For a hyperbola with the equation $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$, the length of the diameter is $2a$.

III. Properties and Characteristics:

1. Symmetry:

• The diameter is the axis of symmetry for a hyperbola. It divides the hyperbola into two symmetrical branches.
2. Foci Relationship:

• The foci of the hyperbola are located on the major axis, and the diameter passes through both foci.
• The sum of the distances from any point on the hyperbola to the foci is constant and equal to the length of the major axis.

Example:

Consider the hyperbola given by the equation:

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

Let's find the diameter of this hyperbola.

1. Identify Key Parameters:

• The given hyperbola is in the form $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$.
• In this case, $a=4$ (semi-major axis) and $b=3$ (semi-minor axis).

2. Calculate the Diameter:

• The length of the diameter ($D$) is twice the length of the semi-major axis ($a$): $D=2a$

• Substitute $a=4$ into the formula: $D=2×4=8$

3. Interpretation:

• The diameter of the given hyperbola is $8$.

4. Verification:

• We can verify this by choosing two points on the hyperbola that lie on a line passing through the center and calculating the distance between them.

• For instance, consider two points: $A\left(4,3\right)$ and $B\left(-4,-3\right)$. These points are symmetric with respect to the center at the origin.

• The distance between these points using the distance formula: $AB=\sqrt{\left(-4-4{\right)}^{2}+\left(-3-3{\right)}^{2}}=\sqrt{64+36}=\sqrt{100}=10$ As expected, $AB$ is equal to the length of the diameter ($D$).