Why Parabolas Matter in JEE
Ever watched a satellite dish or a water fountain curve and wondered how that shape is described in math? That curve is a parabola, and JEE loves to test it.
💡 In Simple Words: A parabola is the set of points that stay equally far from a fixed point (called the focus) and a fixed line (called the directrix). Imagine a flashlight beam that spreads out symmetrically – the bright edge follows a parabola.
What is a Parabola?
In plain language, picture a smooth U‑shaped curve. If you pick any spot on that curve, the distance to the focus and the distance to the directrix are exactly the same. That definition works no matter how the curve is rotated.
Standard Forms of a Parabola
The most common ways JEE asks you to write a parabola are:
- Vertical opening (up or down):
(x-h)^2 = 4a(y-k) - Horizontal opening (right or left):
(y-k)^2 = 4a(x-h)
Here (h,k) is the vertex – the turning point of the curve. The symbol a tells you how “wide” the parabola is and which way it opens. If a is positive, a vertical parabola opens upward; if negative, it opens downward. For a horizontal one, positive a opens right, negative opens left.
Key Properties: Focus, Directrix, Axis
Every parabola has three friends:
- Focus: a single point inside the curve.
- Directrix: a straight line outside the curve.
- Axis: the line that runs through the vertex and focus, acting like a mirror.
Why do they matter? JEE problems often give you one of these pieces and ask you to write the equation, or they give the equation and ask you to locate the focus and directrix.
How to Derive the Equation from Geometry
When you know the focus and directrix, you can build the equation step by step. Below is a quick flowchart that shows the usual order.
Let’s walk through those steps with a concrete example.
Worked Example
Problem: Find the equation of a parabola whose focus is (2,3) and whose directrix is the line y = 1.
Step 1 – Identify given elements: Focus (2,3), directrix y=1. Since the directrix is horizontal, the parabola opens vertically.
Step 2 – Choose orientation: Vertex will lie midway between focus and directrix along the line perpendicular to the directrix. The midpoint in the y‑direction is (1+3)/2 = 2. So vertex is (2,2).
Step 3 – Write standard form: For a vertical parabola opening upward, use (x-h)^2 = 4a(y-k) with (h,k) = (2,2).
Step 4 – Substitute focus/directrix: Distance from vertex to focus = a. Here a = 3-2 = 1. Plugging in, we get (x-2)^2 = 4·1·(y-2) → (x-2)^2 = 4(y-2).
Step 5 – Simplify (optional): Expand if you need the general form: x^2 - 4x + 4 = 4y - 8 → x^2 - 4x - 4y + 12 = 0.
That’s the full answer. You can check quickly: any point on the curve should be equally far from (2,3) and the line y=1.
Quick Summary Table
| Feature | Vertical Parabola | Horizontal Parabola |
|---|---|---|
| Standard form | (x‑h)² = 4a(y‑k) | (y‑k)² = 4a(x‑h) |
| Opens | Up if a>0, Down if a | Right if a>0, Left if a |
| Focus | (h, k+a) | (h+a, k) |
| Directrix | y = k‑a | x = h‑a |
| Axis | Line x = h | Line y = k |
Common Pitfalls to Avoid
- Mixing up
asign – it decides the opening direction. - For a given focus & directrix, forgetting that the vertex is the midpoint between them.
- Skipping the “square” step: remember the distance formula involves squaring, so you often end up with a squared term.
📝 Likely Exam Questions
- Question: Write the equation of a parabola with vertex (‑1, 4) that opens leftward and has focus (‑3, 4).
Answer: Horizontal form(y‑4)² = 4a(x+1). Since focus is 2 units left of vertex, a = –2. Equation:(y‑4)² = –8(x+1). - Question: Find the focus and directrix of the parabola
x² = 12y.
Answer: Compare with(x‑0)² = 4a(y‑0)→ 4a = 12 → a = 3. Focus = (0, a) = (0, 3). Directrix: y = –a = –3. - Question: A parabola passes through (2, 5) and has focus (2, 3). Its axis is vertical. Determine its equation.
Answer: Vertex lies midway: y‑coordinate = (3+?)/2 =? Actually distance from vertex to focus = a. Let vertex be (2, k). Then a = 3‑k. Directrix: y = k‑a = 2k‑3. Use point (2,5): (2‑2)² = 4a(5‑k) → 0 = 4a(5‑k) → 5 = k. So vertex (2,5), a = 3‑5 = –2 (opens downward). Equation: (x‑2)² = –8(y‑5). - Question: Convert the general equation
9x² + 18x + 4y – 5 = 0to standard parabola form and state its focus.
Answer: Complete the square: 9(x²+2x) = 5‑4y → 9[(x+1)²‑1] = 5‑4y → 9(x+1)² = 14‑4y → (x+1)² = (14‑4y)/9 → (x+1)² = –(4/9)(y‑3.5). So a = –1/9·? Actually 4a = –4/9 → a = –1/9. Vertex (‑1, 3.5). Focus = (‑1, 3.5 + a) = (‑1, 3.3889).