Co-ordinate Geometry – JEE Main Maths Formula Sheet | Straight Lines, Circle, Parabola, Ellipse & Hyperbola Formulas

Fundamental Formulas — Distance, Section, Area, Locus, Axes Transformation

Why These Co-ordinate Geometry Foundation Formulas Are Tested in Every JEE Main Session

Distance Formula (from PDF — Co-ordinate Geometry): Distance between P(x₁,y₁) and Q(x₂,y₂): PQ = √[(x₂–x₁)² + (y₂–y₁)²]. Valid for points in any quadrant. Distance from P(x₁,y₁) to origin: OP = √(x₁²+y₁²). Four standard quadrilateral tests: Square (4 sides equal + diagonals equal); Rhombus (4 sides equal, diagonals unequal); Rectangle (opposite sides equal + diagonals equal); Parallelogram (opposite sides equal, diagonals unequal). All four have diagonals bisecting each other.

Section Formula (from PDF — Co-ordinate Geometry): Point R(x,y) dividing P(x₁,y₁) and Q(x₂,y₂):

Internal division in ratio m₁:m₂: x = (m₁x₂+m₂x₁)/(m₁+m₂); y = (m₁y₂+m₂y₁)/(m₁+m₂)

External division in ratio m₁:m₂: x = (m₁x₂–m₂x₁)/(m₁–m₂); y = (m₁y₂–m₂y₁)/(m₁–m₂) [m₁≠m₂]

Midpoint (m₁=m₂): ((x₁+x₂)/2, (y₁+y₂)/2)

For ratio λ:1 division: x = (λx₂+x₁)/(λ+1); positive λ → internal, negative λ → external.

Line Ax+By+C=0 divides PQ in ratio: –(Ax₁+By₁+C)/(Ax₂+By₂+C)

Area of Triangle (from PDF — Co-ordinate Geometry Stair Method): For A(x₁,y₁), B(x₂,y₂), C(x₃,y₃):

Area = (1/2)|x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)| = (1/2)|det[x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1]|

Area of triangle formed by lines a₁x+b₁y+c₁=0, a₂x+b₂y+c₂=0, a₃x+b₃y+c₃=0: Area = Δ²/(2|C₁C₂C₃|) where C₁,C₂,C₃ are cofactors of c₁,c₂,c₃.

Collinearity: A, B, C collinear iff Area = 0 iff determinant = 0.

Area of polygon (Stair Method): A = (1/2)|Σ(xₙyₙ₊₁ – xₙ₊₁yₙ)| (cyclic sum).

Translation and Rotation of Axes (from PDF — Co-ordinate Geometry):

Shifting origin to (h,k): x = X + h, y = Y + k; equivalently X = x–h, Y = y–k.

Rotation through angle θ: X = x cosθ + y sinθ; Y = y cosθ – x sinθ.

Special results: Third vertex of equilateral triangle if two vertices (x₁,y₁), (x₂,y₂) are given: x₃ = (x₁+x₂ ∓ √3(y₁–y₂))/2; y₃ = (y₁+y₂ ± √3(x₁–x₂))/2. Third vertex of right-angle triangle at hypotenuse: x₃ = (x₁+x₂ ± (y₁–y₂))/2; y₃ = (y₁+y₂ ∓ (x₁–x₂))/2. Download the Free PDF for Co-ordinate Geometry for all foundation formula examples.

Straight Lines — All 8 Forms and Key Results

Why All 8 Forms of Equation of a Straight Line Are Core Co-ordinate Geometry JEE Main Formulas

Slope of a line (from PDF — Co-ordinate Geometry): m = tanθ where θ is the inclination (anticlockwise from positive x-axis). Slope of line joining (x₁,y₁) and (x₂,y₂): m = (y₂–y₁)/(x₂–x₁) [x₁≠x₂]. Slope of ax+by+c=0: m = –a/b. Slope of x-axis = 0; y-axis = undefined (∞). Lines equally inclined to both axes: m = ±1.

All 8 Forms of Straight Line Equation (from PDF — Co-ordinate Geometry JEE Main):

(i) General form: ax + by + c = 0. Slope = –a/b; x-intercept = –c/a; y-intercept = –c/b.

(ii) Point-slope form: y – y₁ = m(x – x₁) [slope m, passes through (x₁,y₁)]

(iii) Two-point form: (y–y₁)/(y₂–y₁) = (x–x₁)/(x₂–x₁)

(iv) Slope-intercept form: y = mx + c [slope m, y-intercept c]

(v) Intercept form: x/a + y/b = 1 [x-intercept a, y-intercept b]

(vi) Normal (Perpendicular) form: x cosα + y sinα = p [perpendicular distance p from origin, perpendicular makes angle α with OX]. Reduction from ax+by+c=0: if c<0 → p=–c/√(a²+b²), cosα=a/√(a²+b²), sinα=b/√(a²+b²); if c>0 reverse signs.

(vii) Symmetric (Parametric) form: (x–x₁)/cosθ = (y–y₁)/sinθ = r [point (x₁,y₁), angle θ with x-axis, r = directed distance from (x₁,y₁) to (x,y)]. So x = x₁+r cosθ; y = y₁+r sinθ. Length of segment from (x₁,y₁) to line ax+by+c=0 along this direction: r = –(ax₁+by₁+c)/(a cosθ+b sinθ).

(viii) Lines parallel to axes: y = k (parallel to x-axis); x = k (parallel to y-axis). Equation of x-axis: y=0; y-axis: x=0.

Parallel and perpendicular conditions (Co-ordinate Geometry — JEE Main): Lines with slopes m₁, m₂: parallel iff m₁=m₂; perpendicular iff m₁m₂=–1. Line parallel to ax+by+c=0: ax+by+λ=0. Line perpendicular to ax+by+c=0: bx–ay+λ=0.

Angle between two lines (from PDF — Co-ordinate Geometry): tanθ = |(m₁–m₂)/(1+m₁m₂)|. Two lines making angles ±θ with line y=mx+c through (x₁,y₁): slope m₁ = (m+tanθ)/(1–m·tanθ); m₂ = (m–tanθ)/(1+m·tanθ).

Concurrent lines (from PDF — Co-ordinate Geometry): a₁x+b₁y+c₁=0, a₂x+b₂y+c₂=0, a₃x+b₃y+c₃=0 concurrent iff det[aᵢ bᵢ cᵢ] = 0. Family of lines through intersection of L₁=0 and L₂=0: L₁+λL₂=0. Download the Free PDF for Co-ordinate Geometry for all straight line formula examples.

Distance Formulas, Foot of Perpendicular, Image, Angle Bisectors, and Triangle Centres

Why These Co-ordinate Geometry Point-Line Relations Are High-Value JEE Main Questions

Perpendicular distance from point to line (from PDF — Co-ordinate Geometry): Distance from P(x₁,y₁) to ax+by+c=0: p = |ax₁+by₁+c|/√(a²+b²). Perpendicular from origin to ax+by+c=0: |c|/√(a²+b²). Distance between parallel lines ax+by+c₁=0 and ax+by+c₂=0: |c₁–c₂|/√(a²+b²).

Foot of perpendicular (from PDF — Co-ordinate Geometry): Foot of perpendicular from P(x₁,y₁) to ax+by+c=0: (x–x₁)/a = (y–y₁)/b = –(ax₁+by₁+c)/(a²+b²). So x = x₁ – a(ax₁+by₁+c)/(a²+b²); y = y₁ – b(ax₁+by₁+c)/(a²+b²).

Image (reflection) of point in a line (from PDF — Co-ordinate Geometry): Image of P(x₁,y₁) in ax+by+c=0: (x–x₁)/a = (y–y₁)/b = –2(ax₁+by₁+c)/(a²+b²). So x = x₁ – 2a(ax₁+by₁+c)/(a²+b²); y = y₁ – 2b(ax₁+by₁+c)/(a²+b²). [Note: image formula has 2× the correction, foot formula has 1×]

Angle bisectors (from PDF — Co-ordinate Geometry): Bisectors of angles between a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0 (constants positive): (a₁x+b₁y+c₁)/√(a₁²+b₁²) = ±(a₂x+b₂y+c₂)/√(a₂²+b₂²). '+' gives bisector of angle containing origin. If a₁a₂+b₁b₂>0: '+' sign → obtuse angle bisector; '–' sign → acute angle bisector. If a₁a₂+b₁b₂<0: '+' → acute; '–' → obtuse.

Five Triangle Centres (from PDF — Co-ordinate Geometry JEE Main): For triangle A(x₁,y₁), B(x₂,y₂), C(x₃,y₃), with BC=a, CA=b, AB=c:

(i) Centroid G: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

(ii) Orthocentre H: H = ((x₁tanA+x₂tanB+x₃tanC)/(tanA+tanB+tanC), (y₁tanA+y₂tanB+y₃tanC)/(tanA+tanB+tanC))

(iii) Circumcentre O: O = ((x₁sin2A+x₂sin2B+x₃sin2C)/(sin2A+sin2B+sin2C), (y₁sin2A+y₂sin2B+y₃sin2C)/(sin2A+sin2B+sin2C))

(iv) Incentre I: I = ((ax₁+bx₂+cx₃)/(a+b+c), (ay₁+by₂+cy₃)/(a+b+c)) where a=BC, b=CA, c=AB.

(v) Excentres I₁, I₂, I₃: I₁ = (–ax₁+bx₂+cx₃)/(–a+b+c) etc. (opposite to vertex A, external bisector at A).

Euler Line property (from PDF): O, G, H are collinear (the Euler line) and G divides OH in the ratio 1:2 (OG:GH = 1:2). For right-angled triangle: midpoint of hypotenuse = circumcentre; right-angle vertex = orthocentre. Download the Free PDF for Co-ordinate Geometry for all five centres and Euler line examples for JEE Main.

Standard area formulas (from PDF — Co-ordinate Geometry): Area of parallelogram between 4 lines: |(c₁–d₁)(c₂–d₂)/(a₁b₂–a₂b₁)|. Area of rhombus ax±by±c=0: 2c²/(ab). Area of triangle with sides y=m₁x+c₁, y=m₂x+c₂, y=m₃x+c₃: (1/2)·Σ(cᵢ–cⱼ)²/|(mᵢ–mⱼ)|.

Circle — All Standard Results, Tangent, Chord, Common Tangents, Radical Axis

Why Circle Formulas Are Among the Highest-Density Co-ordinate Geometry Topics in JEE Main

Standard equations of circle (from PDF — Co-ordinate Geometry):

Standard form: (x–h)² + (y–k)² = r² [centre (h,k), radius r]

General form: x² + y² + 2gx + 2fy + c = 0 [centre (–g,–f), radius r = √(g²+f²–c)]. Real circle: g²+f²–c>0; Point circle: g²+f²–c=0; Imaginary: g²+f²–c<0. Circle passes through origin: c=0.

For general second degree ax²+2hxy+by²+2gx+2fy+c=0 to represent a circle: a=b, h=0, g²+f²–ac≥0.

Diametric form: If (x₁,y₁) and (x₂,y₂) are diametrically opposite: (x–x₁)(x–x₂) + (y–y₁)(y–y₂) = 0.

Parametric form: x = h+r cosθ; y = k+r sinθ (0≤θ<2π).

Special cases from PDF: Centre (α,β) touching x-axis: (x–α)²+(y–β)²=β² (radius=|β|). Touching y-axis: (x–α)²+(y–β)²=α² (radius=|α|). Touching both axes: (x±a)²+(y±a)²=a² (four circles, all radii=a).

Intercepts on axes (from PDF — Co-ordinate Geometry): x-intercept of x²+y²+2gx+2fy+c=0: 2√(g²–c). y-intercept: 2√(f²–c). Circle touches x-axis: g²=c; touches y-axis: f²=c; touches both: c=g²=f².

Tangent to circle (from PDF — Co-ordinate Geometry JEE Main): Define S=x²+y²+2gx+2fy+c; S₁=x₁²+y₁²+2gx₁+2fy₁+c; T=xx₁+yy₁+g(x+x₁)+f(y+y₁)+c.

Tangent at (x₁,y₁) to S=0: T=0 [i.e. xx₁+yy₁+g(x+x₁)+f(y+y₁)+c=0]

For x²+y²=a²: tangent at (x₁,y₁) is xx₁+yy₁=a².

Line y=mx+c tangent to x²+y²=a² iff c²=a²(1+m²) → tangent: y=mx±a√(1+m²).

Chord of x²+y²=a² intercepted on y=mx+c: length = 2√(a²–c²/(1+m²))/√(1+m²)... from PDF: 2√[a²(1+m²)–c²]/√(1+m²). Wait — from PDF exactly: length = 2√[a²(1+m²) – c²]/√(1+m²)... = 2√[a² – c²/(1+m²)]. Standard formula: 2√(a²–d²) where d is perpendicular distance from centre.

Point position, tangent length, pair of tangents, chord of contact, chord midpoint (from PDF — Co-ordinate Geometry):

Point (x₁,y₁) inside/on/outside circle S=0: S₁ < 0 / = 0 / > 0.

Length of tangent from (x₁,y₁) to S=0: √S₁ = √(x₁²+y₁²+2gx₁+2fy₁+c).

Pair of tangents from (x₁,y₁): SS₁ = T² (combined equation).

Chord of contact (chord joining points of tangency from P(x₁,y₁)): T=0.

Chord whose midpoint is (x₁,y₁): T=S₁.

Angle between circles, orthogonal circles, radical axis (from PDF — Co-ordinate Geometry):

Angle θ between S₁=0 and S₂=0: cosθ = (2g₁g₂+2f₁f₂–c₁–c₂)/(2r₁r₂).

Orthogonal circles (θ=90°): 2g₁g₂ + 2f₁f₂ = c₁+c₂.

Radical axis of S₁=0 and S₂=0: S₁–S₂=0 (locus of point with equal tangent lengths to both circles).

Common chord of S₁ and S₂: S₁–S₂=0. Common tangent at touching point of S₁ and S₂: S₁–S₂=0.

Common tangents (from PDF — Co-ordinate Geometry JEE Main) — 4 cases:

O₁O₂ > r₁+r₂ → 4 common tangents (2 direct, 2 transverse).

O₁O₂ = r₁+r₂ → 3 common tangents (externally touching).

|r₁–r₂| < O₁O₂ < r₁+r₂ → 2 common tangents.

O₁O₂ = |r₁–r₂| → 1 common tangent (internally touching).

O₁O₂ < |r₁–r₂| → 0 common tangents (one inside the other).

External centre of similitude T₁ divides O₁O₂ externally in r₁:r₂; internal T₂ divides internally in r₁:r₂. Download the Free PDF for Co-ordinate Geometry for all circle results for JEE Main.

Parabola — All 4 Standard Forms, Parameters, Tangents, Normals

Why All 4 Standard Parabola Forms and Parametric Results Are Direct JEE Main Co-ordinate Geometry Questions

Parabola definition (from PDF — Co-ordinate Geometry): Locus of a point equidistant from a fixed point (focus) and a fixed line (directrix). Eccentricity e=1.

All 4 Standard Parabola Forms (from PDF — Co-ordinate Geometry JEE Main):

(i) y² = 4ax (a>0, opens right): Vertex (0,0); Focus S(a,0); Directrix x=–a; Axis y=0. Parametric: x=at², y=2at. Latus rectum: x=a, length=4a, ends (a,±2a).

(ii) y² = –4ax (opens left): Vertex (0,0); Focus (–a,0); Directrix x=a; Axis y=0. Parametric: x=–at², y=2at.

(iii) x² = 4ay (opens up): Vertex (0,0); Focus (0,a); Directrix y=–a; Axis x=0. Parametric: x=2at, y=at².

(iv) x² = –4ay (opens down): Vertex (0,0); Focus (0,–a); Directrix y=a; Axis x=0. Parametric: x=2at, y=–at².

Focal chord (from PDF — Co-ordinate Geometry): If (at₁², 2at₁) and (at₂², 2at₂) are ends of a focal chord of y²=4ax: t₁t₂=–1. Semilatus rectum = HM of focal radii b and c = 2bc/(b+c). Tangents at ends of focal chord are perpendicular and meet on directrix. Sum of slopes of normals drawn from any external point = 0. Sum of ordinates of feet of normals = 0.

Focal distance: For point (x₁,y₁) on y²=4ax: focal distance = x₁+a (= distance from directrix).

Point position for y²=4ax: (x₁,y₁) inside/on/outside: y₁²–4ax₁ <0 / =0 / >0.

Tangent formulas for y²=4ax (from PDF — Co-ordinate Geometry JEE Main):

At (x₁,y₁): yy₁ = 2a(x+x₁) [T=0 form]

At (at², 2at): yt = x + at²

Slope form: y = mx + a/m (tangent condition: c=a/m); point of contact: (a/m², 2a/m).

Pair of tangents from (x₁,y₁): SS₁=T² where S=y²–4ax, S₁=y₁²–4ax₁, T=yy₁–2a(x+x₁).

Chord of contact from (x₁,y₁): T=0 → yy₁=2a(x+x₁).

Chord with midpoint (x₁,y₁): T=S₁.

Two tangents at (at₁²,2at₁) and (at₂²,2at₂) meet at: (at₁t₂, a(t₁+t₂)).

Normal formulas for y²=4ax (from PDF — Co-ordinate Geometry JEE Main):

At (x₁,y₁): y–y₁ = –(y₁/2a)(x–x₁)

At (at²,2at): y = –tx + 2at + at³

Slope form: y = mx – 2am – am³ [normal at (am²,–2am)]. Three normals can be drawn from an external point. Normals at (at₁²,2at₁) and (at₂²,2at₂) meet at: (a(t₁²+t₂²+t₁t₂+2), –at₁t₂(t₁+t₂)). If normal at t₁ meets parabola at t₂: t₂ = –t₁ – 2/t₁. Download the Free PDF for Co-ordinate Geometry for all parabola results for JEE Main.

Ellipse — Complete Comparison Table, Tangents, Normals, Special Results

Why the Ellipse Comparison Table Is the Most Formula-Rich Co-ordinate Geometry Section in JEE Main

Ellipse definition (from PDF — Co-ordinate Geometry): Locus of a point P such that SP/PM = e < 1 (distance from focus / distance from directrix). Physical definition: sum of focal distances PS+PS' = 2a = constant > 2ae (distance between foci).

Complete Comparison Table — Two Standard Ellipse Forms (from PDF — Co-ordinate Geometry):

Horizontal Ellipse: x²/a²+y²/b²=1 (a>b>0)

Centre: (0,0); Major axis: y=0, length 2a; Minor axis: x=0, length 2b

Foci: (±ae, 0); Vertices: (±a, 0) and (0, ±b); Directrices: x=±a/e

Eccentricity: e=√(1–b²/a²) = √(a²–b²)/a; b²=a²(1–e²)

Latus rectum length: 2b²/a; Ends: (±ae, ±b²/a)

Focal radii: SP=a–ex₁; S'P=a+ex₁; SP+S'P=2a

Parametric: x=a cosφ, y=b sinφ (φ = eccentric angle)

Distance between foci: 2ae; Distance between directrices: 2a/e

Tangents at vertices: x=±a

Vertical Ellipse: x²/b²+y²/a²=1 (a>b>0)

Centre: (0,0); Major axis: x=0, length 2a; Minor axis: y=0, length 2b

Foci: (0, ±ae); Vertices: (0, ±a) and (±b, 0); Directrices: y=±a/e

Same eccentricity e; Latus rectum: 2b²/a

Focal radii: SP=a–ey₁; S'P=a+ey₁

Parametric: x=b cosφ, y=a sinφ

Auxiliary circle (from PDF — Co-ordinate Geometry): Circle on major axis as diameter: x²+y²=a². If Q=(acosφ, asinφ) is on auxiliary circle, then P=(acosφ, bsinφ) is on ellipse. PN/QN = b/a where N is foot from P to major axis.

Tangent to Ellipse x²/a²+y²/b²=1 (3 forms — Co-ordinate Geometry JEE Main):

At (x₁,y₁): T=0 → xx₁/a² + yy₁/b² = 1

At (acosφ, bsinφ): (x cosφ)/a + (y sinφ)/b = 1

Slope form: y=mx±√(a²m²+b²); point of contact: (∓a²m/√(a²m²+b²), ±b²/√(a²m²+b²))

Condition: y=mx+c tangent iff c²=a²m²+b².

Chord of contact from (x₁,y₁): T=0 → xx₁/a²+yy₁/b²=1.

Director circle (from PDF): locus of perpendicular tangent intersection: x²+y²=a²+b².

Chord with midpoint (x₁,y₁): T=S₁.

Chord joining (acosα,bsinα) and (acosβ,bsinβ): (x/a)cos((α+β)/2) + (y/b)sin((α+β)/2) = cos((α–β)/2).

Two tangents at (acosφ₁,bsinφ₁) and (acosφ₂,bsinφ₂) meet at: (acos((φ₁+φ₂)/2)/cos((φ₁–φ₂)/2), bsin((φ₁+φ₂)/2)/cos((φ₁–φ₂)/2)).

Normals to Ellipse (from PDF — Co-ordinate Geometry):

At (x₁,y₁): (a²x–b²y)/(x–y) = a²–b² or a²x/x₁ – b²y/y₁ = a²–b²

At (acosφ, bsinφ): ax secφ – by cosecφ = a²–b²

Slope form: y=mx–(a²–b²)m/√(a²+b²m²)... from PDF: y=mx–m(a²–b²)/√(a²+b²m²)... exact: y = mx – (a²–b²)m/(√(a²+b²m²)). Four normals from any external point; sum of eccentric angles of feet = odd multiple of π. Download the Free PDF for Co-ordinate Geometry for all ellipse results for JEE Main.

Hyperbola — Standard Form, Rectangular Hyperbola xy=c², Tangents, Normals

Why Hyperbola and Rectangular Hyperbola Are the Most Distinctive Co-ordinate Geometry Conic JEE Main Topics

Hyperbola definition (from PDF — Co-ordinate Geometry): Locus of point P with PS/PM = e > 1. |SP–S'P| = 2a (difference of focal distances). The standard hyperbola: x²/a²–y²/b²=1 where b²=a²(e²–1).

Standard Hyperbola x²/a²–y²/b²=1 (from PDF — Co-ordinate Geometry JEE Main):

Centre: (0,0); Vertices: (±a, 0); Foci: S(ae,0), S'(–ae,0)

Directrices: x=±a/e; Eccentricity: e=√(1+b²/a²)=√(a²+b²)/a; b²=a²(e²–1)

Transverse axis: y=0, length=2a; Conjugate axis: x=0, length=2b (BB' where CB=CB'=b)

Latus rectum: x=±ae, length=2b²/a

Focal radii: SP=ex₁–a; S'P=ex₁+a; |SP–S'P|=2a

Parametric: x=asecφ, y=btanφ (0≤φ<2π)

Auxiliary circle: x²+y²=a²

Tangent to Hyperbola (from PDF — Co-ordinate Geometry):

At (x₁,y₁): T=0 → xx₁/a²–yy₁/b²=1

At (asecφ, btanφ): (x secφ)/a – (y tanφ)/b = 1

Slope form: y=mx±√(a²m²–b²); condition: c²=a²m²–b²

Two tangents at (asecφ₁,btanφ₁) and (asecφ₂,btanφ₂) meet at: (acos((φ₁–φ₂)/2)/cos((φ₁+φ₂)/2), bsin((φ₁+φ₂)/2)/cos((φ₁+φ₂)/2))... from PDF directly.

Normals to Hyperbola (from PDF — Co-ordinate Geometry):

At (x₁,y₁): a²x/x₁ + b²y/y₁ = a²+b²

At (asecφ, btanφ): ax cosφ + by cotφ = a²+b²

Standard results: Chord of contact from (x₁,y₁): T=0. Chord midpoint (x₁,y₁): T=S₁. Pair of tangents: SS₁=T². Director circle: x²+y²=a²–b².

Point position for hyperbola x²/a²–y²/b²=1: (x₁,y₁) inside/on/outside: x₁²/a²–y₁²/b²–1 >0 / =0 / <0. [Note: OPPOSITE sign convention to ellipse — the conic-positive region is inside hyperbola]

Rectangular (Equilateral) Hyperbola xy=c² (from PDF — Co-ordinate Geometry JEE Main):

This is x²–y²=a² rotated 45°, with c²=a²/2. Eccentricity e=√2.

Centre (0,0); asymptotes: y=±x (i.e. the coordinate axes after rotation → coordinate axes ARE the asymptotes).

Transverse axis: y=x; Conjugate axis: y=–x.

Parametric: x=ct, y=c/t (parameter t).

Tangent at (ct, c/t): x + yt² = 2ct → divide by t: x/t + yt = 2c.

Normal at (ct, c/t): xt³ – yt – ct⁴ + c = 0 → from PDF: tx – yt³ – c(t⁴–1) = 0... exact from PDF: normal at (ct,c/t): xt³ – yt = c(t⁴–1) (from: slope of normal at (ct,c/t) = t², normal: y–c/t = t²(x–ct)). Download the Free PDF for Co-ordinate Geometry for all hyperbola and rectangular hyperbola results for JEE Main.

General Conic — Discriminant Conditions and Classification

Why the General Second Degree Equation Classification Is a Direct JEE Main Co-ordinate Geometry Question

General second degree equation (from PDF — Co-ordinate Geometry): ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a conic section. Define:

Discriminant Δ = abc + 2fgh – af² – bg² – ch²

Classification of conic (from PDF — Co-ordinate Geometry JEE Main):

(i) Δ = 0 → pair of straight lines (degenerate conic)

(ii) Δ ≠ 0, h=0, a=b (a≠0), g²+f²–ac≥0 → circle

(iii) Δ ≠ 0, h²=ab → parabola (e=1)

(iv) Δ ≠ 0, h²<ab → ellipse (e<1) [includes circle as special case]

(v) Δ ≠ 0, h²>ab → hyperbola (e>1)

Note: For a circle (h=0, a=b): h²=0<a²=ab only if a=b, consistent with h²<ab for a=b>0 only when h=0. Rectangular hyperbola: h²=ab and a+b=0.

Conic section as cone sections (from PDF — Co-ordinate Geometry): Cutting a right circular cone with a plane: parallel to base → circle; parallel to one generator → parabola; neither parallel to generator nor to base → ellipse; parallel to axis → hyperbola; through vertex → pair of lines; at vertex parallel to base → point.

Eccentricity summary (from PDF — Co-ordinate Geometry): Point: e not defined. Circle: e=0. Parabola: e=1. Ellipse: 0<e<1. Hyperbola: e>1. Rectangular hyperbola: e=√2. Download the Free PDF for Co-ordinate Geometry for complete general conic classification and all conic parameter tables for JEE Main.