JEE Advanced Mensuration: Surface Areas & Volumes — practice questions — page 2

Focus on the **negative $y$-axis** ⇒ the parabola opens **downward** with axis along the $y$-axis. Standard form: $x^2 = -4ay$. $|a| = 3 \Rightarrow x^2 = -12y$. **Trap.** Option D ($y^2 = 12x$) puts the focus on the positive $x$-axis option C on the negative $x$-axis. Always pick the standard form from the *axis* the focus lies on (vertex at origin pins the rest).

**Setup.** Axis along positive $x$-axis with vertex at origin ⇒ standard form $y^2 = 4ax$. **Plug in the known point.** $(4, 6)$ on the curve: $6^2 = 4a \cdot 4 \Rightarrow 36 = 16a \Rightarrow a = \dfrac{9}{4}$. **Substitute back.** $y^2 = 4 \cdot \dfrac{9}{4} \cdot x = 9x$. Quick verification: $(4, 6)$ ⇒ $36 = 9 \times 4 = 36$ ✓. **Distractor check.** Option B ($y^2 = 4x$) corresponds to $a = 1$ — would pass through $(4, 4)$, not $(4, 6)$. Option D ($a = 9$) overshoots: $(4, ?) \Rightarrow y^2 = 144$, $y = 12$.

**Identify $a$.** $y^2 = 4ax$ with $4a = 16 \Rightarrow a = 4$. **Locate the directrix.** For $y^2 = 4ax$ the directrix is the vertical line $x = -a$ (on the opposite side of the vertex from the focus). So $x = -4$. **Geometric sanity check.** Pick any point on $y^2 = 16x$, say $(1, 4)$. Distance to focus $(4, 0)$: $\sqrt{9 + 16} = 5$. Distance to directrix $x = -4$: horizontal $|1 - (-4)| = 5$. ✓ Equal — confirming the directrix. **Trap options.** $x = 4$ (B) is the focus's $x$-coordinate, not the directrix. Options C and D ($y = \pm 4$) put a *horizontal* directrix, which is wrong for an axis-on-$x$ parabola.