Problem Statement
For the hyperbola $9x^2 - 16y^2 = 144$, find:
(i) Length of transverse and conjugate axes
(ii) Coordinates of the vertices and foci
(iii) Eccentricity
(iv) Length of the latus rectum.
Verified Solution & Marking Scheme
Convert to standard form
Dividing by $144$:
$$\frac{9x^2}{144} - \frac{16y^2}{144} = 1 \implies \frac{x^2}{16} - \frac{y^2}{9} = 1$$
Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get $a^2 = 16 \implies a = 4$, and $b^2 = 9 \implies b = 3$.
Axes and Vertices
Transverse axis length $= 2a = 2(4) = 8$.
Conjugate axis length $= 2b = 2(3) = 6$.
Vertices: $(\pm a, 0) = (\pm 4, 0)$.
Eccentricity and Foci
$$c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$
$$e = \frac{c}{a} = \frac{5}{4}$$
Foci coordinates: $(\pm c, 0) = (\pm 5, 0)$.
Length of Latus Rectum
$$\text{L.R.} = \frac{2b^2}{a} = \frac{2(9)}{4} = \frac{9}{2} = 4.5\text{ units}$$