Ever wondered why the path of a basketball shot looks like a perfect smile? That curve is a parabola, and it shows up everywhere in JEE coordinate geometry.
A parabola is the set of points that are equally distant from a fixed point (called the focus) and a straight line (called the directrix). Think of it like a garden hose: the water (points) spreads out so that every droplet is the same distance from the nozzle (focus) and the garden wall (directrix).
What is a Parabola?
In plain language, a parabola is a U‑shaped curve that opens either up, down, left, or right. It’s the shape you get when you slice a right circular cone parallel to its side. In JEE problems, you’ll mostly meet the “vertical” parabola that opens upward or downward.
Focus‑Directrix Definition
The focus is a single point inside the curve. The directrix is a line that never touches the curve. Every point on the parabola keeps the same distance to both. If you pick a point P on the curve, draw a line to the focus (PF) and drop a perpendicular to the directrix (PD). PF = PD.
Standard Equation
When the vertex (the turning point) sits at the origin (0,0) and the parabola opens upward, the equation is
y² = 4ax (if it opens right) or x² = 4ay (if it opens up). The letter a is the distance from the vertex to the focus. If a is positive, the curve opens right/up; if negative, left/down.
Vertex Form
Most JEE questions give the vertex at (h, k). Then the equation becomes
(y‑k)² = 4a(x‑h) for a sideways opening, or (x‑h)² = 4a(y‑k) for a vertical opening. This form lets you read the vertex directly.
Key Parameters at a Glance
| Parameter | Meaning | How to Find |
|---|---|---|
| Vertex (h,k) | Turn‑around point of the curve | Given or from completing the square |
| Focus | Point (h+a, k) for horizontal, (h, k+a) for vertical | Use the value of a |
| Directrix | Line x = h‑a (horizontal) or y = k‑a (vertical) | Again, a tells you the offset |
| Axis of symmetry | Line passing through vertex and focus | x = h (horizontal) or y = k (vertical) |
| Latus rectum | Chord through the focus, perpendicular to the axis; its length = 4|a| | 4 times the absolute value of a |
How to Write the Equation from Given Elements
JEE loves to hide the parabola’s equation behind a few clues. Here’s a quick recipe:
- Identify the vertex (h, k). If it’s not given, find it by symmetry or by completing the square.
- Determine whether the parabola opens horizontally or vertically – look at the given focus or directrix.
- Calculate a = distance between vertex and focus (or vertex and directrix). Remember the sign: a > 0 for opening towards the focus.
- Plug h, k, and a into the appropriate vertex form.
Worked Example
Problem: Find the equation of the parabola whose vertex is (2,‑1) and whose focus is (2,3).
Solution:
- Vertex (h, k) = (2,‑1). Since the focus shares the same x‑coordinate, the axis is vertical and the parabola opens upward.
- Distance a = focus y‑coordinate – vertex y‑coordinate = 3 – (‑1) = 4.
- Use the vertical vertex form: (x‑h)² = 4a(y‑k). Plug in h = 2, k = ‑1, a = 4:
- (x‑2)² = 4·4 (y + 1) → (x‑2)² = 16(y + 1).
Check: The focus should be (2, k + a) = (2, ‑1 + 4) = (2,3). It matches, so the equation is correct.
Quick JEE Tricks for Parabolas
- Remember the “4a” pattern – it appears in every standard form. Spotting it saves time.
- If the question gives the length of the latus rectum, directly write a = length/4.
- When the parabola is rotated (rare in JEE), use the general second‑degree equation and check the discriminant B²‑4AC = 0.
- For tangent problems, the slope form y = mx + c can be substituted into the parabola equation; the condition for a single intersection gives a quadratic with discriminant zero.
- Always verify the sign of a by checking where the focus lies relative to the vertex.
📝 Likely Exam Questions
- Question: The parabola y² = 12x is shifted right by 3 units and up by 2 units. Write its new equation.
Answer: Replace x by (x‑3): y² = 12(x‑3). Then shift up: (y‑2)² = 12(x‑3). - Question: Find the focus and directrix of the parabola (x‑1)² = 8(y + 2).
Answer: Compare with (x‑h)² = 4a(y‑k). Here h=1, k=‑2, 4a=8 → a=2. Focus: (h, k+a) = (1, 0). Directrix: y = k‑a = ‑4. - Question: A parabola has vertex (‑3,4) and latus rectum length 20. Write its equation.
Answer: Length = 4|a| → |a| = 5. Since latus rectum is positive, a = 5. Parabola opens upward (a>0). Equation: (x+3)² = 4·5 (y‑4) → (x+3)² = 20(y‑4). - Question: Determine the slope(s) of the tangent(s) to y = x² at points where the tangent passes through (0,‑4).
Answer: Tangent at (t, t²) has slope m = 2t. Equation: y‑t² = 2t(x‑t). Plug (0,‑4): ‑4‑t² = 2t(‑t) → ‑4‑t² = ‑2t² → ‑4 = ‑t² → t² = 4 → t = ±2. Slopes: m = 2t = ±4. - Question: Find the area enclosed by the parabola y = 4x – x² and the x‑axis.
Answer: Roots: 4x‑x² = 0 → x(4‑x)=0 → x=0,4. Area = ∫₀⁴ (4x‑x²) dx = [2x²‑(1/3)x³]₀⁴ = (2·16‑(1/3)·64) = 32‑21.33… = 10.67 units² (≈32/3).