Indian Navy MR Mathematics — practice by topic

Eccentricity scale across the conic family: | Conic | $e$ | |---|---| | Circle | $0$ | | Ellipse | $0 < e < 1$ | | Parabola | $e = 1$ | | **Hyperbola** | **$e > 1$** | For a hyperbola, $c^2 = a^2 + b^2 \Rightarrow c > a \Rightarrow e = c/a > 1$.

Use the identity $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\dfrac{a + b}{1 - ab}\right)$, valid when $ab < 1$. With $a = 1/2$, $b = 1/3$: $ab = 1/6 < 1$, so the identity applies cleanly. $\dfrac{a + b}{1 - ab} = \dfrac{1/2 + 1/3}{1 - 1/6} = \dfrac{5/6}{5/6} = 1$. So the sum equals $\tan^{-1}(1) = \dfrac{\pi}{4}$.

Direction ratios of the line joining two points are the differences in coordinates: $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$. Here: $(4 - 1, 6 - 2, 9 - 3) = (3, 4, 6)$. Direction ratios are not unique. $(6, 8, 12)$ or $(1.5, 2, 3)$ would also be valid since multiplying by any non-zero scalar preserves the direction. The standard answer just uses the raw differences. Direction cosines (the normalised version) are $(3, 4, 6) / \sqrt{9 + 16 + 36} = (3, 4, 6) / \sqrt{61}$.