The directrix of the parabola $4 y^{2} + 12 x - 12 y + 39 = 0$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — $x = - \frac{7}{4}$

Detailed Solution

Complete the square in $y$ : $4 (y - \frac{3}{2})^{2} - 9 + 12 x + 39 = 0$ , so $4 (y - \frac{3}{2})^{2} = - 12 x - 30$ .

$(y - \frac{3}{2})^{2} = - 3 (x + \frac{5}{2})$ . The vertex is $(- \frac{5}{2}, \frac{3}{2})$ , the parabola opens to the left, and $4 a = 3$ , so $a = \frac{3}{4}$ .

For a left-opening parabola the directrix lies $a$ units to the right of the vertex: $x = - \frac{5}{2} + \frac{3}{4} = - \frac{7}{4}$ .

✓ Check

The distance from the vertex to this line is

\frac{3}{4} = a

(see figure).

⚠ Common trap

Subtracting

a

instead gives

- \frac{13}{4}

, which is the focus, not the directrix.

Quick tip

$(y - k)^{2} = - 4 a (x - h) \Rightarrow$ directrix $x = h + a$

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Parabola — Preparation

This question tests your understanding of Parabola, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A parabola is the set of points equidistant from a fixed point (the focus) and a fixed line (the directrix), so its eccentricity is exactly 1. The standard form y^2 = 4ax has its vertex at the origin, focus at (a, 0), directrix x = -a, and latus rectum of length 4a. The other standard forms open left, up or down. Exam questions test identifying the focus, directrix, axis and latus rectum, the parametric point (at^2, 2at), focal distances, and equations of tangents and normals. A key result is that the line y = mx + c touches y^2 = 4ax when c = a/m. Drawing a quick sketch of which way the parabola opens prevents most sign errors. Practise the parabola questions below, each with a step-by-step solution.

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The directrix of the parabola $4 y^{2} + 12 x - 12 y + 39 = 0$ is

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