# Eccentricity – Hyperbola

## 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.