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Parabola questions in JEE Advanced use a small set of results again and again: the focal chord condition, the tangent and normal at a point, and a few facts that follow. You can memorise them, but each takes two lines to derive from the parametric point, and deriving them once protects you against misremembering a sign. This post derives each result and checks all of them on one numerical example.
Where is this on the syllabus?
The JEE (Advanced) 2026 brochure lists, for the parabola, ellipse and hyperbola, "parametric equations, equations of tangent and normal". The JEE (Main) 2026 syllabus lists the conic sections "in standard forms", without a separate item on tangents and normals. The current NCERT Class 11 chapter on conic sections defines the parabola y² = 4ax with focus (a, 0), directrix x = −a and latus rectum of length 4a; it does not cover tangents, normals or focal chords. So treat this post as JEE Advanced preparation, and check the documents for your exam year.
Why use the point (at², 2at)?
Because it is on the curve for every value of t: (2at)² = 4a²t² = 4a(at²). One letter now describes any point, instead of x and y tied together by an equation. Two more facts make it powerful:
Focal distance. By the definition of a parabola, the distance of a point from the focus equals its distance from the directrix x = −a. For (at², 2at), that is at² + a = a(1 + t²).
Chord slope. The chord joining t₁ and t₂ has slope 2a(t₁ − t₂)/[a(t₁² − t₂²)] = 2/(t₁ + t₂).
What is the focal chord condition?
Put the focus (a, 0) on the chord through t₁ and t₂:
0 − 2at₁ = [2/(t₁ + t₂)](a − at₁²).
Multiply out: −2at₁² − 2at₁t₂ = 2a − 2at₁². The t₁² terms cancel, leaving t₁t₂ = −1.
Two results follow quickly. With t₂ = −1/t, the focal chord has length a(1 + t²) + a(1 + 1/t²) = a(t + 1/t)². Since (t + 1/t)² is at least 4, the shortest focal chord is 4a, the latus rectum. And 1/SP + 1/SQ = 1/a for the two focal distances SP and SQ.
What are the tangent and normal at t?
Tangent. The tangent to y² = 4ax at (x₁, y₁) is yy₁ = 2a(x + x₁). At (at², 2at), this becomes ty = x + at². Its slope is 1/t. Writing m = 1/t gives the slope form y = mx + a/m, which is the condition for y = mx + c to touch the parabola: c = a/m.
Normal. The normal is perpendicular to the tangent, so its slope is −t. Through (at², 2at): y = −tx + 2at + at³.
Worked example: check every result on y² = 8x
Here 4a = 8, so a = 2. The focus is (2, 0) and the directrix is x = −2.
The point. Take t = 2: P = (at², 2at) = (8, 8). Check: 8² = 64 = 8 × 8.
The focal chord. The other end has t = −1/2: Q = (2 × 1/4, 2 × 2 × (−1/2)) = (0.5, −2). Check: (−2)² = 4 = 8 × 0.5.
Its length. SP = a + x = 2 + 8 = 10 and SQ = 2 + 0.5 = 2.5, so PQ = 12.5. The formula gives a(t + 1/t)² = 2 × (2.5)² = 12.5. And 1/10 + 1/2.5 = 0.1 + 0.4 = 0.5 = 1/a.
The tangents. At t = 2: 2y = x + 8. At t = −1/2: −y/2 = x + 0.5, so y = −2x − 1. Solving together: 2(−2x − 1) = x + 8 gives x = −2, y = 3. The point (−2, 3) is on the directrix, and the slopes 1/2 and −2 multiply to −1.
The normal at P. y = −2x + 2(2)(2) + 2(8) = −2x + 24. Check: at x = 8, y = 8.
Which mistakes cost the most marks?
Using the slope form c = a/m for a parabola that is not y² = 4ax, such as x² = 4ay. Rederive for the new orientation.
Writing the focal chord condition as t₁t₂ = 1.
Mixing up the slopes: the tangent at t has slope 1/t and the normal has slope −t.
Forgetting that a is a quarter of the coefficient: y² = 8x has a = 2, not 8.
Not sketching. A quick figure shows at once whether a point should be above or below the axis.
How should you practise?
Draw the figure first, every time. Larkin and Simon (1987) argued that a diagram displays information that is only implicit in equations and would otherwise have to be computed. In parabola problems, that is often the position of the focus or which side of the axis a point is on.
Derive from scratch (once a week, 15 minutes). Focal chord condition, tangent, normal and the tangent intersection point, from (at², 2at) alone.
One parabola, every result. Pick a and t, and check all the results above numerically, as in the worked example.
Mixed conics. Mix parabola questions with ellipse and hyperbola ones, as in our five-step method for conic sections. Our guide to the ellipse for JEE uses the same approach with the eccentric angle.
If the geometry is hard to picture, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Today, repeat the worked example with y² = 12x and t = 3, and confirm that the two tangents meet on x = −3.
