The eccentricity of a hyperbola is a measure of how much the hyperbola deviates from being circular. It is denoted by e.
For a hyperbola, where:
- a = distance from the center to a vertex
- c = distance from the center to a focus
Focus-Directrix Formula
For a hyperbola having,
- PF = distance between a point on a hyperbola and its focus,
- PD = distance between the same point and the directrix of the hyperbola.
e=\frac{PF}{PD}
Formula for the Standard Equation
For a hyperbola having the following equation,
x2/a2 - y2/b2 = 1
the formula for eccentricity is
e = \sqrt{1 + \frac{b^2}{a^2}}
- a = length of transverse axis of the hyperbola,
- b = length of conjugate axis of the hyperbola.
Derivation
Using the equation of the hyperbola,
x2/a2 - y2/b2 = 1
Let the coordinates for the foci for the hyperbola be F(c,0) and F'(-c,0). As per the definition of a hyperbola, for a point on a hyperbola P(x,y).
PF' - PF = 2a
By using the formula for distance between two points, we get,
√((x+c)2+y2) - √((x-c)2+y2) = 2a
√((x+c)2+y2) = 2a + √((x-c)2+y2)
On squaring both sides, we get,
(x+c)2 + y2 = 4a2 + (x-c)2 + y2 + 4a√((x-c)2+y2)
On further simplification and rearrangement of terms, we get,
x2/a2 - y2/(c2-a2) = 1
On comparing with the equation of hyperbola, we get,
c2 - a2 = b2
Now, as per the definition of a hyperbola, If we take the point on the hyperbola as the vertex of the hyperbola. Then, we get after substituting the required distances in above equation,
e = √(a2+b2)/a
or, e = √(1 + b2/a2)
Thus, we have derived the expression for the eccentricity of the hyperbola in terms of the lengths of its transverse axis and conjugate axis.
Solved Examples
Example 1: Find the value for eccentricity of a hyperbola represented by the following equation: x2/144 - y2/36 = 1.
Comparing the given equation with standard equation of hyperbola, we get,
a2 = 144 and b2 = 36, giving a = 12 and b = 6.
We know that, eccentricity (e) is given by,
e = √(1 + b2/a2)
⇒ e = √(1 + 36/144) = √(5/4) = √5/2
Thus, eccentricity of given hyperbola comes out to be 1.12 approximately.
Example 2: Find the value for eccentricity of a hyperbola represented as x2/16 - y2/64 = 1.
For the given hyperbola, we have, a2 = 16 and b2 = 64
Formula for eccentricity of the hyperbola, e = √(1 + b2/a2)
⇒ e = √(1 + 64/16)
⇒ e = √(1 + 4) = √5
⇒ e = √5
Hence, eccentricity of given hyperbola comes out to be √5 which equals to 2.24 approximately.
Practice Problems
P1: What is the eccentricity value for the given hyperbola: x2 - y2 = 5?
P2: Find the eccentricity for the hyperbola represented as x2/81 - y2/121 = 1.
P3: What is the value of eccentricity for a hyperbola given by x2 - y2 = 1?
P4: A hyperbola lies along the X-axis. The lengths of its transverse and conjugate axes are 8 units and 2 units, respectively. Find the value of eccentricity for this hyperbola.
P5: The equation of a hyperbola is written as x/16 - y2/b2 = 1. It has eccentricity value as √ 3. Find the value of b.
