Indian Navy 10+2 B.Tech Cadet Mathematics — practice by topic

**Geometric reading.** $|z - 2|$ is the distance from the point $z$ to $(2, 0)$. $|z + 2|$ is the distance to $(-2, 0)$. Equating them means $z$ is equidistant from the two points. **Locus.** Set of points equidistant from $(2, 0)$ and $(-2, 0)$ is the **perpendicular bisector** of the segment joining them — the $y$-axis, i.e. $\mathrm{Re}(z) = 0$. **Algebraic confirmation.** Let $z = x + iy$. Squaring: $(x-2)^2 + y^2 = (x+2)^2 + y^2 \Rightarrow -4x = 4x \Rightarrow x = 0$. **Why option C is tempting.** A symmetric-looking radius-2 hint, but $|z - 2|$ is *not* the same as $|z| - 2$ it's a distance to a *point*, not a centred circle equation.

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

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 $(3, 4, 6) / \sqrt{9 + 16 + 36} = (3, 4, 6) / \sqrt{61}$.