1. Chapter at a Glance
A conic section is the curve obtained by cutting a double cone with a plane. Depending on the angle of the cut we get a circle, parabola, ellipse or hyperbola. Every conic can be described as the locus of a point whose distance from a fixed point (the focus) is e times its distance from a fixed line (the directrix); e is the eccentricity. JEE Main asks one or two questions, mostly on the parabola and ellipse.
- Focus, directrix, eccentricity and latus rectum of standard and shifted conics.
- Tangent to a conic with a given slope (condition of tangency).
- Focal chord properties of the parabola.
- Finding the equation of an ellipse or hyperbola from its foci and eccentricity.
| Eccentricity | Conic |
|---|---|
| e = 0 | Circle |
| 0 < e < 1 | Ellipse |
| e = 1 | Parabola |
| e > 1 | Hyperbola (e = √2 for a rectangular hyperbola) |
2. Parabola
Standard form y² = 4ax (a > 0): the set of points equidistant from the focus S(a, 0) and the directrix x = −a.
| Feature | y² = 4ax | x² = 4ay |
|---|---|---|
| Focus | (a, 0) | (0, a) |
| Directrix | x = −a | y = −a |
| Axis | y = 0 | x = 0 |
| Latus rectum | 4a | 4a |
| Parametric point | (at², 2at) | (2at, at²) |
- Tangent at (x1, y1): yy1 = 2a(x + x1); at t: ty = x + at².
- Tangent with slope m: y = mx + a/m (condition c = a/m).
- Focal distance of (x1, y1) = x1 + a.
- Focal chord: if the ends are t1 and t2, then t1t2 = −1. The semi-latus rectum 2a is the harmonic mean of the two focal segments: 1/SP + 1/SQ = 1/a.
- Tangents at the ends of a focal chord meet at right angles on the directrix.
3. Ellipse
Standard form x²/a² + y²/b² = 1 with a > b > 0: the set of points whose distances from two foci add up to 2a.
- Major axis 2a (along x), minor axis 2b; latus rectum 2b²/a.
- For any point P on the ellipse, SP + S′P = 2a.
- Parametric point: (a cos θ, b sin θ).
- Tangent at (x1, y1): xx1/a² + yy1/b² = 1. Tangent with slope m: y = mx ± √(a²m² + b²).
- Director circle (perpendicular tangents): x² + y² = a² + b². Area = πab.
- If b > a, the major axis is along the y-axis: then a² = b²(1 − e²) and the foci are (0, ±be).
4. Hyperbola
Standard form x²/a² − y²/b² = 1: the set of points whose distances from two foci differ by 2a.
- Transverse axis 2a, conjugate axis 2b; latus rectum 2b²/a.
- |SP − S′P| = 2a for every point P on it.
- Asymptotes: y = ±(b/a)x. The curve approaches these lines but never meets them.
- Tangent with slope m: y = mx ± √(a²m² − b²).
- Rectangular hyperbola (a = b): x² − y² = a², e = √2; also xy = c², with parametric point (ct, c/t).
- The conjugate hyperbola x²/a² − y²/b² = −1 has eccentricity e′ with 1/e² + 1/e′² = 1.
5. Formula Master Table
| Feature | Parabola y² = 4ax | Ellipse x²/a² + y²/b² = 1 | Hyperbola x²/a² − y²/b² = 1 |
|---|---|---|---|
| Eccentricity | 1 | √(1 − b²/a²) | √(1 + b²/a²) |
| Foci | (a, 0) | (±ae, 0) | (±ae, 0) |
| Directrix | x = −a | x = ±a/e | x = ±a/e |
| Latus rectum | 4a | 2b²/a | 2b²/a |
| Tangent, slope m | y = mx + a/m | y = mx ± √(a²m² + b²) | y = mx ± √(a²m² − b²) |
| Tangent at (x1, y1) | yy1 = 2a(x + x1) | xx1/a² + yy1/b² = 1 | xx1/a² − yy1/b² = 1 |
| Parametric point | (at², 2at) | (a cos θ, b sin θ) | (a sec θ, b tan θ) |
| Focal distances | x + a | a ± ex | |ex ± a| |
6. Mind Map
- SP = e·PM
- e decides the type
- Focus (a, 0), LR 4a
- c = a/m
- Focal chord t1t2 = −1
- b² = a²(1 − e²)
- SP + S′P = 2a
- Area πab
- b² = a²(e² − 1)
- |SP − S′P| = 2a
- Asymptotes y = ±bx/a
7. Tips and Tricks
- For a shifted conic like (y − k)² = 4a(x − h), find everything for y² = 4ax and then shift by (h, k).
- Always complete the square first when the equation has linear terms, e.g. y² − 8y = 4x.
- For an ellipse, "distance between foci = 2ae" and "distance between directrices = 2a/e" — their ratio gives e².
- If a line is a tangent to both a parabola and a circle, use both tangency conditions simultaneously.
- Assuming a is under x² in an ellipse even when the larger denominator is under y².
- Using b² = a²(1 − e²) for a hyperbola — the correct relation is b² = a²(e² − 1).
- Forgetting that the latus rectum of the parabola y² = 4ax is 4a, not 2a.
8. Important One-liners
- The vertex of a parabola is the midpoint of the focus and the foot of the directrix.
- Every ray parallel to the axis of a parabola reflects through its focus (used in dish antennas).
- The ellipse becomes a circle as e → 0.
- The length of the latus rectum of an ellipse is 2b²/a.
- The asymptotes of a rectangular hyperbola are perpendicular.
- The eccentricities of a hyperbola and its conjugate satisfy 1/e² + 1/e′² = 1.
- The product of perpendiculars from the foci to any tangent of an ellipse equals b².
9. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
The focus of the parabola y² = 12x is:
- (0, 3)
- (3, 0)
- (12, 0)
- (−3, 0)
Show solution
4a = 12 ⇒ a = 3, so the focus is (3, 0).
Answer: (B)
The eccentricity of the ellipse x²/25 + y²/16 = 1 is:
- 4/5
- 3/5
- 3/4
- 5/3
Show solution
a² = 25, b² = 16: e = √(1 − 16/25) = √(9/25) = 3/5.
Answer: (B)
The length of the latus rectum of the ellipse x²/25 + y²/9 = 1 is:
- 18/5
- 10/3
- 9/5
- 6
Show solution
Latus rectum = 2b²/a = 2 × 9/5 = 18/5.
Answer: (A)
The foci of the hyperbola x²/16 − y²/9 = 1 are:
- (±4, 0)
- (±5, 0)
- (0, ±5)
- (±√7, 0)
Show solution
e = √(1 + 9/16) = 5/4, so ae = 4 × 5/4 = 5. Foci (±5, 0).
(√7 would be the answer for an ellipse with these numbers — a common slip.)
Answer: (B)
The equation of the tangent to the parabola y² = 8x with slope 1 is:
- y = x + 1
- y = x + 2
- y = x + 4
- y = x − 2
Show solution
4a = 8 ⇒ a = 2. Tangent with slope m: y = mx + a/m = x + 2.
Answer: (B)
An ellipse has foci (±3, 0) and major axis of length 10. Its equation is:
- x²/25 + y²/16 = 1
- x²/25 + y²/9 = 1
- x²/100 + y²/91 = 1
- x²/16 + y²/25 = 1
Show solution
2a = 10 ⇒ a = 5; ae = 3 ⇒ a²e² = 9. b² = a² − a²e² = 25 − 9 = 16.
Answer: (A)
Find the length of the latus rectum of the parabola y² − 8y = 4x.
Show solution
Complete the square: (y − 4)² = 4x + 16 = 4(x + 4). This is of the form Y² = 4aX with 4a = 4.
Latus rectum = 4a = 4.
Answer: 4
If SP and SQ are the segments of a focal chord of the parabola y² = 8x (S is the focus), then 1/SP + 1/SQ equals:
- 1/2
- 1/4
- 2
- 1
Show solution
For y² = 4ax, 1/SP + 1/SQ = 1/a (the semi-latus rectum 2a is the harmonic mean of SP and SQ).
Here a = 2, so 1/SP + 1/SQ = 1/2. Check with the latus rectum itself: SP = SQ = 2a = 4 ⇒ 1/4 + 1/4 = 1/2.
Answer: (A)
10. Quick Revision Summary
- e = 0 circle, e < 1 ellipse, e = 1 parabola, e > 1 hyperbola.
- Parabola y² = 4ax: focus (a, 0), directrix x = −a, LR 4a, tangent y = mx + a/m.
- Ellipse: b² = a²(1 − e²), LR 2b²/a, SP + S′P = 2a, tangent c² = a²m² + b².
- Hyperbola: b² = a²(e² − 1), asymptotes y = ±bx/a, tangent c² = a²m² − b².
- Complete the square for shifted conics; check which denominator is larger.