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