JEE PREP | APRIL 2026
Last Updated: April 2026
Coordinate Geometry is one of the most predictable and high-scoring topics in JEE Main Mathematics. With 5–7 questions in JEE Main 2027, mastering Straight Lines, Circles, and Parabola gives you a guaranteed score boost. This guide covers every standard formula, JEE-level problem patterns, and common mistakes to avoid.
JEE Main 2022–2026 average: 5–7 questions from Coordinate Geometry. Very high predictability — same formula types repeated year after year. Straight Lines (2 questions), Circles (2 questions), Parabola (1–2 questions) is the typical split. Speed and formula accuracy are key. Avoid common errors in distance from a line and tangent conditions.
• Distance between points (x1,y1) and (x2,y2): d = √[(x2-x1)² + (y2-y1)²]
• Distance from point (x1,y1) to line ax+by+c=0: |ax1+by1+c|/√(a²+b²)
• Circle: (x-h)²+(y-k)²=r²; tangent condition: c²=a²(1+m²)
• Parabola y²=4ax: Focus (a,0), Directrix x=-a, Latus rectum length=4a
• Two lines are perpendicular if m1×m2=-1; parallel if m1=m2
Section 1: Straight Lines
Basic Formulas
Distance and Section Formulas
- Distance formula: d = √[(x₂-x₁)² + (y₂-y₁)²]
- Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2)
- Section formula (internal): P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
- Area of triangle: A = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
- Collinearity condition: Area = 0
Slope and Equation of Line
| Form | Equation | When to Use |
|---|---|---|
| Slope-intercept | y = mx + c | Slope m and y-intercept c known |
| Point-slope | y – y₁ = m(x – x₁) | One point and slope known |
| Two-point form | (y-y₁)/(y₂-y₁) = (x-x₁)/(x₂-x₁) | Two points known |
| Intercept form | x/a + y/b = 1 | x-intercept a and y-intercept b known |
| Normal form | x·cosα + y·sinα = p | p = perpendicular distance from origin, α = angle of normal |
| General form | ax + by + c = 0 | Universal form |
Angle Between Two Lines
If m₁ and m₂ are slopes of two lines:
tan θ = |m₁ – m₂| / |1 + m₁m₂|
- Parallel lines: m₁ = m₂
- Perpendicular lines: m₁ × m₂ = -1
Distance of Point from Line
Distance of point (x₁, y₁) from line ax + by + c = 0:
d = |ax₁ + by₁ + c| / √(a² + b²)
Example: Distance of (3, 4) from 3x – 4y + 5 = 0:
d = |3(3) – 4(4) + 5| / √(9 + 16) = |9 – 16 + 5| / 5 = |-2| / 5 = 2/5 → Wait, = 2 (since |9-16+5|=|-2|=2, √25=5, so d=2/5? No — d = 2/5 if we use this formula. Let’s verify: 9-16+5 = -2, |-2| = 2, √(9+16) = 5, so d = 2/5. But if we use (3,4) from 3x-4y+5=0 correctly: d = |9-16+5|/5 = |-2|/5 = 2/5. Actual answer = 2/5 = 0.4. In MCQ context often rounded to 2.)
Correct calculation: |3(3)-4(4)+5|/√(3²+4²) = |9-16+5|/5 = 2/5
Family of Lines
The equation of any line through the intersection of L₁: a₁x+b₁y+c₁=0 and L₂: a₂x+b₂y+c₂=0 can be written as:
L₁ + λL₂ = 0, i.e., (a₁x+b₁y+c₁) + λ(a₂x+b₂y+c₂) = 0
Key JEE Patterns — Straight Lines
- Angle bisectors of two lines: use |(a₁x+b₁y+c₁)/√(a₁²+b₁²)| = |(a₂x+b₂y+c₂)/√(a₂²+b₂²)|
- Concurrent lines: three lines are concurrent if their determinant = 0
- Foot of perpendicular from point to line: use (x-x₁)/a = (y-y₁)/b = -(ax₁+by₁+c)/(a²+b²)
Section 2: Circles
General Equation of Circle
Standard form: (x-h)² + (y-k)² = r², center (h,k), radius r
General form: x² + y² + 2gx + 2fy + c = 0, center (-g,-f), radius = √(g²+f²-c)
Tangent to a Circle
1. Tangent at point (x₁, y₁) on circle x²+y²=a²: xx₁ + yy₁ = a²
2. Tangent y=mx+c to circle x²+y²=a²: condition is c² = a²(1+m²)
or equivalently c = ±a√(1+m²)
3. Length of tangent from external point (x₁,y₁) to circle x²+y²+2gx+2fy+c=0:
L = √(x₁²+y₁²+2gx₁+2fy₁+c)
Chord of Contact
Equation of chord of contact (line joining the two points where tangents from external point touch the circle):
From point (x₁, y₁) to circle x²+y²=a²: xx₁ + yy₁ = a² (same form as tangent at a point!)
Position of Point/Line with Respect to Circle
- Point (x₁,y₁) lies inside circle x²+y²=a² if x₁²+y₁² < a²
- Point lies on circle if x₁²+y₁² = a²
- Point lies outside circle if x₁²+y₁² > a²
Two Circle Relationships
| Condition (d = distance between centers) | Relationship |
|---|---|
| d > r₁ + r₂ | External to each other (4 common tangents) |
| d = r₁ + r₂ | Externally tangent (3 common tangents) |
| |r₁ – r₂| < d < r₁ + r₂ | Two intersection points (2 common tangents) |
| d = |r₁ – r₂| | Internally tangent (1 common tangent) |
| d < |r₁ - r₂| | One circle inside the other (0 common tangents) |
Section 3: Parabola
Standard Forms of Parabola
| Equation | Vertex | Focus | Directrix | Axis |
|---|---|---|---|---|
| y² = 4ax | (0,0) | (a,0) | x = -a | x-axis (y=0) |
| y² = -4ax | (0,0) | (-a,0) | x = a | x-axis, opens left |
| x² = 4ay | (0,0) | (0,a) | y = -a | y-axis |
| (y-k)² = 4a(x-h) | (h,k) | (h+a, k) | x = h-a | y = k |
Parametric Form
For parabola y²=4ax: Any point can be written as (at², 2at) where t is the parameter.
Tangent to Parabola y²=4ax
- At point (x₁,y₁): yy₁ = 2a(x+x₁)
- In slope form (m): y = mx + a/m (condition: c = a/m for tangency)
- Parametric form at parameter t: ty = x + at²
Properties of Focal Chord
A focal chord passes through the focus (a,0). For a chord with end points t₁ and t₂ on y²=4ax:
- t₁ × t₂ = -1 (for focal chord)
- Length of focal chord = a(t₁ – t₂)² = a(t + 1/t)² (for parameter t)
- Length of latus rectum = 4a (the shortest focal chord, perpendicular to axis)
“Focus = (a,0) — The coefficient of x divided by 4 gives ‘a'”
For y²=8x: 4a=8, so a=2. Focus = (2,0), Directrix: x=-2, Latus rectum = 8
For y²=12x: 4a=12, so a=3. Focus = (3,0), Directrix: x=-3
Quick check: Point (x,y) on y²=4ax? Substitute and verify y²=4ax.
10 Solved JEE-Level Problems
Q1. Find the equation of line through (1,2) and (3,6).
Sol: Slope m = (6-2)/(3-1) = 4/2 = 2. Using y-2 = 2(x-1): y = 2x. Check: (3,6) → 6 = 2×3 ✓
Q2. Find distance from (3,4) to line 3x-4y+5=0.
Sol: d = |3(3)-4(4)+5|/√(9+16) = |9-16+5|/5 = |-2|/5 = 2/5
Q3. Find the tangent from origin to circle x²+y²-6x-4y+8=0.
Sol: Length = √(0+0-0-0+8) = √8 = 2√2
Q4. Find focus of parabola y²=16x.
Sol: 4a=16, a=4. Focus = (4,0).
Q5. Find the acute angle between lines y=2x+3 and y=3x+1.
Sol: m₁=2, m₂=3. tan θ = |2-3|/|1+6| = 1/7. θ = arctan(1/7)
Q6. If line y=2x+c is tangent to circle x²+y²=5, find c.
Sol: c²=a²(1+m²)=5(1+4)=25. c=±5.
Q7. A point on parabola y²=8x has parameter t=2. Find coordinates.
Sol: a=2. Point = (at², 2at) = (2×4, 2×2×2) = (8, 8). Verify: 8²=64=8×8 ✓
Q8. Find area of triangle with vertices A(0,0), B(3,0), C(0,4).
Sol: Area = ½|0(0-4)+3(4-0)+0(0-0)| = ½|12| = 6 sq units.
Q9. Two circles have centers (0,0) and (5,0) with radii 2 and 2. Are they intersecting?
Sol: d=5, r₁+r₂=4, |r₁-r₂|=0. Since d>r₁+r₂, circles are external to each other.
Q10. If the chord of contact from point P(h,k) to circle x²+y²=9 passes through (2,3), find the locus of P.
Sol: Chord of contact: hx+ky=9. This passes through (2,3): 2h+3k=9. Locus: 2x+3y=9.
Common Mistakes to Avoid
1. Distance formula sign error: Always use |ax₁+by₁+c|/√(a²+b²) — the absolute value is mandatory
2. Tangent condition: c²=a²(1+m²) NOT c=a(1+m) — square the equation properly
3. Parabola ‘a’ value: For y²=4ax, a is the coefficient of x divided by 4, NOT 4 itself
4. Chord of contact vs tangent at point: Both have equation xx₁+yy₁=a² but contexts differ
5. Circle center: For x²+y²+2gx+2fy+c=0, center is (-g,-f), NOT (g,f)
Practice MCQs — Coordinate Geometry
Test your JEE preparation with these 10 MCQs on Straight Lines, Circles and Parabola:
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Frequently Asked Questions
What is the formula for distance from a point to a line?
The perpendicular distance from a point (x₁, y₁) to a line ax + by + c = 0 is given by: d = |ax₁ + by₁ + c| / √(a² + b²). The absolute value in the numerator is essential — without it you get a signed distance which can be negative. For example, distance of (3,4) from 3x-4y+5=0: d = |9-16+5|/√(9+16) = 2/5. This formula is one of the most commonly tested in JEE Main.
How do you find the focus of a parabola in JEE?
For the standard parabola y² = 4ax, the focus is at (a, 0). To find ‘a’, divide the coefficient of x by 4. For example: y²=8x → 4a=8 → a=2 → Focus=(2,0). For y²=12x → a=3 → Focus=(3,0). For x²=4ay (vertical axis), focus is at (0,a). For a shifted parabola (y-k)²=4a(x-h), the focus is at (h+a, k). Always identify the vertex first, then shift the standard focus accordingly.
What is the condition for a line to be tangent to a circle?
For the line y = mx + c to be tangent to circle x² + y² = a², the condition is c² = a²(1 + m²), which can also be written as c = ±a√(1+m²). Geometrically, this means the perpendicular distance from the center (0,0) to the line equals the radius: a = |c|/√(1+m²), which gives |c| = a√(1+m²), hence c² = a²(1+m²). A common JEE error is writing c = a(1+m) — always square the terms properly.
How many questions come from Coordinate Geometry in JEE Main?
JEE Main typically has 5–7 questions from Coordinate Geometry. The distribution from 2020–2026 is approximately: Straight Lines (1–2 questions), Circles (2–3 questions), Parabola (1–2 questions), with occasional questions on Ellipse and Hyperbola. This makes it one of the most question-dense areas in JEE Mathematics. The chapter is highly formula-driven and predictable — the same formula types (tangent conditions, distance formulas, focus/directrix) appear repeatedly. Focus on speed and formula accuracy.