The coordinates of the focus of the parabola $4 y^{2} + 12 x - 20 y + 67 = 0$ are

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — $(- \frac{17}{4}, \frac{5}{2})$

Detailed Solution

$4 (y^{2} - 5 y) = - 12 x - 67$ , so $4 (y - \frac{5}{2})^{2} - 25 = - 12 x - 67$ .

$(y - \frac{5}{2})^{2} = - 3 x - \frac{21}{2} = - 3 (x + \frac{7}{2})$ : vertex $(- \frac{7}{2}, \frac{5}{2})$ , opening to the left, with $4 a = 3$ , so $a = \frac{3}{4}$ .

Focus $= (- \frac{7}{2} - \frac{3}{4}, \frac{5}{2}) = (- \frac{17}{4}, \frac{5}{2})$ .

✓ Check

The point

(- 6.5, 5.5)

is on the parabola (

3^{2} = 9 = - 3 (- 3)

); its distance to

(- 4.25, 2.5)

is 3.75, equal to its distance from the directrix

x = - 2.75

⚠ Common trap

(B) is the source key; it doubles the

x

-value and halves the

y

-value of the true focus.

Quick tip

Complete the square: $(y - k)^{2} = - 4 a (x - h) \Rightarrow$ focus $(h - a, k) = (- \frac{17}{4}, \frac{5}{2})$

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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 coordinates of the focus of the parabola $4 y^{2} + 12 x - 20 y + 67 = 0$ are

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The coordinates of the focus of the parabola 4y2+12x−20y+67=0 are

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The coordinates of the focus of the parabola $4 y^{2} + 12 x - 20 y + 67 = 0$ are