Coordinate geometry JEE questions: formulas, patterns and the mistakes that cost marks

Straight lines

Slope, intercepts and the distance from a point to a line cover most line questions. The angle between two lines is tan θ = |(m₁ − m₂)/(1 + m₁m₂)|, which also gives the perpendicularity condition m₁m₂ = −1 when the denominator vanishes.

Circles in general form

From x² + y² + 2gx + 2fy + c = 0, the centre is (−g, −f) and the radius is √(g² + f² − c). If that expression is negative the circle is imaginary, which is a favourite way to make a question look harder than it is.

Tangency as an algebraic condition

A line touches a curve when substitution gives a quadratic with zero discriminant. For x² + y² = a² this reduces to c² = a²(1 + m²); each conic has its own version worth memorising.

The parabola

For y² = 4ax the focus is (a, 0), the directrix is x = −a and the latus rectum is 4a. The parametric point (at², 2at) turns chord and normal questions into single-variable algebra.

The ellipse

For x²/a² + y²/b² = 1 with a > b, e² = 1 − b²/a², foci lie at (±ae, 0) and the latus rectum is 2b²/a. If the larger denominator sits under y², the major axis is vertical and the roles swap.

The hyperbola

For x²/a² − y²/b² = 1, e² = 1 + b²/a², so e > 1 always. The asymptotes y = ±(b/a)x are the fastest check on whether you have written the equation the right way round.

Distance from a point to a line

d = |ax₁ + by₁ + c| / √(a² + b²)

Write the line as ax + by + c = 0 first.

Angle between two lines

tan θ = |(m₁ − m₂)/(1 + m₁m₂)|

Perpendicular when m₁m₂ = −1; parallel when m₁ = m₂.

Circle, general form

x² + y² + 2gx + 2fy + c = 0 · centre (−g, −f) · r = √(g² + f² − c)

Real circle only if g² + f² − c > 0.

Tangent to a circle

y = mx ± a√(1 + m²) touches x² + y² = a²

Equivalent to the condition c² = a²(1 + m²).

Parabola

y² = 4ax · focus (a, 0) · directrix x = −a · latus rectum 4a

Parametric point (at², 2at).

Tangent to a parabola

y = mx + a/m touches y² = 4ax

Point of contact (a/m², 2a/m).

Ellipse

x²/a² + y²/b² = 1 (a > b) · e² = 1 − b²/a² · foci (±ae, 0) · LR = 2b²/a

Check which denominator is larger before using this.

Tangent to an ellipse

y = mx ± √(a²m² + b²)

Slope form; the point form is xx₁/a² + yy₁/b² = 1.

Hyperbola

x²/a² − y²/b² = 1 · e² = 1 + b²/a² · asymptotes y = ±(b/a)x

Eccentricity is always greater than 1.

Tangent to a hyperbola

y = mx ± √(a²m² − b²)

Requires a²m² > b² for a real tangent.

Circle · 2 marks · easy

Find the centre and radius of the circle x² + y² − 4x + 6y − 12 = 0.

1. Compare with x² + y² + 2gx + 2fy + c = 0: 2g = −4 gives g = −2, and 2f = 6 gives f = 3, with c = −12.
2. Centre = (−g, −f) = (2, −3).
3. Radius = √(g² + f² − c) = √(4 + 9 + 12) = √25 = 5.
4. Check by completing the square: (x − 2)² + (y + 3)² = 25.

Answer: Centre (2, −3), radius 5.

Parabola · 3 marks · moderate

Find the equation of the tangent of slope 2 to the parabola y² = 8x, and its point of contact.

1. Compare with y² = 4ax: 4a = 8, so a = 2.
2. The slope-form tangent is y = mx + a/m, so with m = 2: y = 2x + 2/2 = 2x + 1.
3. Point of contact is (a/m², 2a/m) = (2/4, 4/2) = (0.5, 2).
4. Verify: substituting y = 2x + 1 into y² = 8x gives 4x² − 4x + 1 = (2x − 1)² = 0, a repeated root at x = 0.5 — exactly one intersection, so the line is a tangent.

Answer: y = 2x + 1, touching the parabola at (0.5, 2).

Ellipse · 3 marks · easy

For the ellipse x²/25 + y²/9 = 1, find the eccentricity, the foci and the length of the latus rectum.

1. The larger denominator is under x², so a² = 25, b² = 9, and the major axis is horizontal.
2. e² = 1 − b²/a² = 1 − 9/25 = 16/25, so e = 4/5.
3. Foci are at (±ae, 0) = (±5 × 4/5, 0) = (±4, 0).
4. Latus rectum = 2b²/a = 2(9)/5 = 3.6.

Answer: e = 4/5; foci (±4, 0); latus rectum 3.6 units.