Question

Find the centre, eccentricity, foci and directions of the hyperbola 

16x² – 9y² + 32x + 36y – 164 = 0

Answer 1

Given: 16x² – 9y² + 32x + 36y – 164 = 0 

To find: center, eccentricity(e), coordinates of the foci f(m, n), equation of directrix. 

16x² – 9y² + 32x + 36y – 164 = 0 

⇒ 16x² + 32x + 16 – 9y² + 36y – 36 – 16 + 36 – 164 = 0 

⇒ 16(x² + 2x + 1) – 9(y² – 4y + 4) – 16 + 36 – 164 = 0 

⇒ 16(x² + 2x + 1) – 9(y² – 4y + 4) – 144 = 0 

⇒ 16(x + 1)² – 9(y – 2)² = 144

Here, center of the hyperbola is (-1, 2) 

Let x + 1 = X and y – 2 = Y

\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)

Formula used: 

For hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1;\)

Eccentricity(e) is given by,

Foci are given by (±ae, 0) 

The equation of directrix are x = \(\pm \frac{a}{e}\)

Length of latus rectum is \(\frac{2b^2}{a}\)

Here, a = 3 and b = 4

⇒ X = ±5 and Y = 0 

⇒ x + 1 = ±5 and y – 2 = 0 

⇒ x = ±5 – 1 and y = 2 

So, Foci: (±5 – 1, 2) 

Equation of directrix are:

\(\Rightarrow\) 5x - 4 = 0 and 5x + 14 = 0