Eccentricity – Hyperbola
The ratio of the distance from center to focus to distance from center to vertex is called eccentricity.
\(e=\frac{Dis\tan ce\ from\ centre\ to\ focus}{Dis\tan ce\ from\ centre\ to\ vertex}=\frac{c}{a}\).
\({{e}^{2}}=\frac{{{c}^{2}}}{{{a}^{2}}}\).
\({{e}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}=1+\frac{{{b}^{2}}}{{{a}^{2}}}\).
a²e² = a² + b²
Therefore, the equation of hyperbola in terms of eccentricity can be written as
\(\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}({{e}^{2}}-1)}=1\).
The coordinates of the foci are (±ae, 0)
Two hyperbolas are said to be similar if they have the same value of eccentricity
Examples 1: Find the eccentricity of the following hyperbola.
(i) 16y² -9x² = 144
(ii) 5x² – 4y² + 20x + 8y = 4
Solutions:
(i) Given that 16y² -9x² = 144
\(\frac{16{{y}^{2}}}{144}-\frac{9{{x}^{2}}}{144}=1\).
\(\frac{{{y}^{2}}}{144/16}-\frac{{{x}^{2}}}{144/9}=1\).
\(\frac{{{y}^{2}}}{9}-\frac{{{x}^{2}}}{16}=1\) … (1)
\(\frac{{{y}^{2}}}{{{a}^{2}}}-\frac{{{x}^{2}}}{{{b}^{2}}}=1\) … (2)
Compare (1) and (2)
a² = 16 ⇒ a = 4
b² = 9 ⇒ b = 3
eccentricity (e) = a²e² = a² + b²
16e² = 16 + 9
16e² = 25
e² = 25/16 ⇒ e = 5/4
(ii) Given that 5x² – 4y² + 20x + 8y = 4
⇒ 5(x² + 4x) – 4(y² – 2y) = 4
5(x² + 4x -4 + 4) – 4(y² – 2y + 1 – 1) = 4
5(x² + 4x + 4) – 4(y² – 2y + 1) = 4 – 4 + 20
5(x² + 4x + 4) – 4(y² – 2y + 1) = 20
5(x + 2)²- 4(y – 2)² = 20
5(x + 2)² / 20 – 4(y – 2)²/ 20 = 1
(x + 2)²/4 – (y – 2)²/ 5 = 1
a² = 4 ⇒ a = 2
b² = 5
eccentricity (e) = a²e² = a² + b²
= 4 x e² = 4 + 5
e² = 9/4
e = 3/2.