Coordinate Geometry Formulas For JEE 2026
Coordinate Geometry (2D) is an important topic in JEE Mathematics and is known for its high weightage, with around 3–4 questions asked every year. This chapter covers key areas like straight lines, circles, parabolas, ellipses, and hyperbolas. It helps students understand how algebra and geometry come together to solve problems involving points, lines, and curves on a coordinate plane.
To perform well in this topic, students should be clear with formulas such as the distance formula, section formula, equations of lines, and standard equations of conics, along with tangent conditions and their properties. Since many questions are directly based on these formulas, regular practice and revision are very important. For quick revision, students can also use a well-organized JEE Mains Mathematics Formula PDF to go through important formulas and concepts easily.
Distance and Section Formulas
Distance Between Two Points
The distance between $A(x_1, y_1)$ and $B(x_2, y_2)$:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$Special case: Distance from origin to $(x, y)$: $d = \sqrt{x^2 + y^2}$
Section Formula
The point dividing the join of $A(x_1, y_1)$ and $B(x_2, y_2)$:
Internal division in ratio $m : n$:
$$P = \left(\frac{mx_2 + nx_1}{m + n},\; \frac{my_2 + ny_1}{m + n}\right)$$External division in ratio $m : n$:
$$P = \left(\frac{mx_2 - nx_1}{m - n},\; \frac{my_2 - ny_1}{m - n}\right)$$Midpoint (ratio $1:1$):
$$M = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right)$$Worked Example
Find the point dividing $A(2, 3)$ and $B(8, -1)$ internally in ratio $2:1$.
$P = \left(\dfrac{2 \cdot 8 + 1 \cdot 2}{3},\; \dfrac{2 \cdot (-1) + 1 \cdot 3}{3}\right) = \boldsymbol{\left(6, \dfrac{1}{3}\right)}$
Area of a Triangle
For vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$:
$$\text{Area} = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$$If the area is zero, the three points are collinear.
Worked Example
Find the area of the triangle with vertices $A(1, 2)$, $B(4, 6)$, $C(7, 2)$.
Area $= \dfrac{1}{2}|1(6-2) + 4(2-2) + 7(2-6)| = \dfrac{1}{2}|4 + 0 - 28| = \textbf{12 square units}$
Straight Lines
Slope
The slope of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. If the line makes angle $\theta$ with the positive $x$-axis, then $m = \tan\theta$.
Forms of a Straight Line
| Form | Equation | When to Use |
|---|---|---|
| Slope-intercept | $y = mx + c$ | Slope $m$ and $y$-intercept $c$ known |
| Point-slope | $y - y_1 = m(x - x_1)$ | Slope $m$ and one point known |
| Two-point | $\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1}$ | Two points known |
| Intercept | $\dfrac{x}{a} + \dfrac{y}{b} = 1$ | $x$-intercept $a$, $y$-intercept $b$ |
| Normal | $x\cos\alpha + y\sin\alpha = p$ | Perpendicular distance $p$, angle $\alpha$ |
| General | $ax + by + c = 0$ | Standard form |
Worked Example
Find the equation of the line passing through $(2, 3)$ with slope $4$.
Using point-slope form: $y - 3 = 4(x - 2) \Rightarrow \boldsymbol{y = 4x - 5}$
Angle Between Two Lines
If two lines have slopes $m_1$ and $m_2$, the acute angle $\theta$ between them:
$$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$$- Parallel lines: $m_1 = m_2$ (slopes are equal)
- Perpendicular lines: $m_1 \cdot m_2 = -1$ (product of slopes is $-1$)
Tip: For lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$: they are parallel if $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$ and perpendicular if $a_1a_2 + b_1b_2 = 0$.
Distance from a Point to a Line
Distance from point $(x_1, y_1)$ to line $ax + by + c = 0$:
$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$Distance Between Two Parallel Lines
Distance between $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$:
$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$Worked Example
Find the distance from $(3, -4)$ to the line $3x - 4y + 5 = 0$.
$d = \dfrac{|3(3) - 4(-4) + 5|}{\sqrt{9 + 16}} = \dfrac{|9 + 16 + 5|}{5} = \dfrac{30}{5} = \textbf{6}$
Family of Lines
Family of Lines
If $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$, then:
$$(a_1x + b_1y + c_1) + \lambda(a_2x + b_2y + c_2) = 0$$represents all lines through the intersection of $L_1$ and $L_2$ (except $L_2$ itself).
Worked Example
Find the line through the intersection of $x + y - 1 = 0$ and $2x - y + 3 = 0$ that passes through the origin.
Family: $(x + y - 1) + \lambda(2x - y + 3) = 0$
Passes through $(0, 0)$: $-1 + 3\lambda = 0 \Rightarrow \lambda = \dfrac{1}{3}$
$\boldsymbol{5x + 2y = 0}$
Tip: The family-of-lines approach avoids the need to first find the intersection point. It is especially useful when the intersection coordinates are messy fractions.
Circle
Standard Form of a Circle
Circle with centre $(h, k)$ and radius $r$:
$$(x - h)^2 + (y - k)^2 = r^2$$Centre at origin: $x^2 + y^2 = r^2$
General Form of a Circle
$$x^2 + y^2 + 2gx + 2fy + c = 0$$- Centre $= (-g, -f)$
- Radius $= \sqrt{g^2 + f^2 - c}$ (exists only if $g^2 + f^2 - c > 0$)
Worked Example
Find the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$.
$2g = -6 \Rightarrow g = -3$, $2f = 4 \Rightarrow f = 2$, $c = -12$
Centre $= (3, -2)$, Radius $= \sqrt{9 + 4 + 12} = \sqrt{25} = \textbf{5}$
Tangent to a Circle
At point $(x_1, y_1)$ on $x^2 + y^2 = r^2$:
$$xx_1 + yy_1 = r^2$$At point $(x_1, y_1)$ on $x^2 + y^2 + 2gx + 2fy + c = 0$:
$$xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$$Tangent of slope $m$ to $x^2 + y^2 = r^2$:
$$y = mx \pm r\sqrt{1 + m^2}$$Condition for Tangency
Line $y = mx + c$ is tangent to $x^2 + y^2 = r^2$ if and only if:
$$c^2 = r^2(1 + m^2)$$Length of Tangent
From external point $(x_1, y_1)$ to circle $x^2 + y^2 + 2gx + 2fy + c = 0$:
$$L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$$Worked Example
Find the length of tangent from $(6, 8)$ to $x^2 + y^2 = 25$.
$L = \sqrt{36 + 64 - 25} = \sqrt{75} = \boldsymbol{5\sqrt{3}}$
Parabola
Parabola
The locus of a point that is equidistant from a fixed point (focus) and a fixed line (directrix).
Standard Forms of Parabola
| Equation | Opens | Focus | Directrix | Latus Rectum |
|---|---|---|---|---|
| $y^2 = 4ax$ | Right | $(a, 0)$ | $x = -a$ | $4a$ |
| $y^2 = -4ax$ | Left | $(-a, 0)$ | $x = a$ | $4a$ |
| $x^2 = 4ay$ | Up | $(0, a)$ | $y = -a$ | $4a$ |
| $x^2 = -4ay$ | Down | $(0, -a)$ | $y = a$ | $4a$ |
Vertex is at the origin for all standard forms. The parameter $a > 0$.
Worked Example
Find the focus, directrix, and latus rectum of $y^2 = 12x$.
$4a = 12 \Rightarrow a = 3$. Focus $= (3, 0)$, Directrix: $x = -3$, Latus Rectum $= 12$
Tangent to a Parabola $y^2 = 4ax$
- At point $(x_1, y_1)$: $yy_1 = 2a(x + x_1)$
- At parameter $t$ (point $(at^2, 2at)$): $ty = x + at^2$
- Slope form: $y = mx + \dfrac{a}{m}$ (tangent of slope $m$, $m \neq 0$)
- Condition for tangency: Line $y = mx + c$ is tangent if $c = \dfrac{a}{m}$
Worked Example
Find the equation of tangent to $y^2 = 8x$ with slope $2$.
$4a = 8 \Rightarrow a = 2$. Using slope form: $y = 2x + \dfrac{2}{2} = \boldsymbol{2x + 1}$
Tip: For the parabola $y^2 = 4ax$, the parametric form is $x = at^2$, $y = 2at$. Using parameter $t$ often simplifies calculations involving tangents, normals, and chords.
Ellipse
Ellipse
The locus of a point such that the sum of its distances from two fixed points (foci) is constant: $PF_1 + PF_2 = 2a$.
Eccentricity of Ellipse
$e = \dfrac{c}{a}$ where $c^2 = a^2 - b^2$. For an ellipse, $0 < e < 1$. If $e = 0$, it is a circle.
Standard Ellipse ($a > b > 0$)
Horizontal major axis: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ ($a > b$)
- Centre: $(0, 0)$
- Foci: $(\pm c, 0)$ where $c = ae = \sqrt{a^2 - b^2}$
- Vertices: $(\pm a, 0)$
- Ends of minor axis: $(0, \pm b)$
- Eccentricity: $e = \sqrt{1 - \dfrac{b^2}{a^2}}$
- Length of latus rectum: $\dfrac{2b^2}{a}$
Vertical major axis: $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ ($a > b$). Foci at $(0, \pm c)$, vertices at $(0, \pm a)$.
Worked Example
Find the foci, eccentricity, and latus rectum of $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.
$a = 5$, $b = 3$. $c = \sqrt{25 - 9} = 4$.
Foci: $(\pm 4, 0)$, $e = \dfrac{4}{5}$, Latus rectum $= \dfrac{18}{5}$
Tangent to an Ellipse
At point $(x_1, y_1)$ on $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$:
$$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$$Slope form: $y = mx \pm \sqrt{a^2m^2 + b^2}$
Condition for tangency: $y = mx + c$ is tangent if $c^2 = a^2m^2 + b^2$.
Hyperbola
Hyperbola
The locus of a point such that the absolute difference of its distances from two fixed points (foci) is constant: $|PF_1 - PF_2| = 2a$.
Standard Hyperbola
Horizontal transverse axis: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$
- Centre: $(0, 0)$
- Foci: $(\pm c, 0)$ where $c = \sqrt{a^2 + b^2}$
- Vertices: $(\pm a, 0)$
- Eccentricity: $e = \dfrac{c}{a} = \sqrt{1 + \dfrac{b^2}{a^2}}$ ($e > 1$ always)
- Asymptotes: $y = \pm\dfrac{b}{a}\,x$
- Length of latus rectum: $\dfrac{2b^2}{a}$
Vertical transverse axis: $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$. Foci at $(0, \pm c)$, vertices at $(0, \pm a)$, asymptotes $y = \pm\dfrac{a}{b}\,x$.
Worked Example
Find the foci, eccentricity, and asymptotes of $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$.
$a = 4$, $b = 3$. $c = \sqrt{16 + 9} = 5$.
Foci: $(\pm 5, 0)$, $e = \dfrac{5}{4}$, Asymptotes: $y = \pm\dfrac{3}{4}x$
Rectangular Hyperbola
When $a = b$: $x^2 - y^2 = a^2$.
- Asymptotes: $y = \pm x$ (perpendicular to each other)
- Eccentricity: $e = \sqrt{2}$
- Rotated form: $xy = c^2$ (asymptotes along the axes)
Tangent to a Hyperbola
At point $(x_1, y_1)$ on $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$:
$$\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$$Slope form: $y = mx \pm \sqrt{a^2m^2 - b^2}$ (tangent exists if $a^2m^2 - b^2 > 0$)
Condition for tangency: $y = mx + c$ is tangent if $c^2 = a^2m^2 - b^2$.
Eccentricity Comparison
Eccentricity of Conics
| Conic | Eccentricity $e$ | Relation |
|---|---|---|
| Circle | $e = 0$ | $a = b$, special case of ellipse |
| Ellipse | $0 < e < 1$ | $c^2 = a^2 - b^2$ |
| Parabola | $e = 1$ | Focus–directrix definition |
| Hyperbola | $e > 1$ | $c^2 = a^2 + b^2$ |
Important Properties of Conics
Key Properties — Quick Reference
- Parabola: The tangent at any point makes equal angles with the focal chord and the axis (reflection property). The foot of the perpendicular from the focus to any tangent lies on the tangent at the vertex.
- Ellipse: $PF_1 + PF_2 = 2a$ for any point $P$. A ray from one focus reflects off the ellipse toward the other focus. The product of perpendicular distances from the foci to any tangent is $b^2$.
- Hyperbola: $|PF_1 - PF_2| = 2a$ for any point $P$. The product of perpendicular distances from the foci to any tangent is $b^2$. The asymptotes pass through the centre.
- Focal chord: For parabola $y^2 = 4ax$, if one end of a focal chord has parameter $t$, the other end has parameter $-1/t$.
- Director circle: Locus of point from which two perpendicular tangents are drawn. For ellipse: $x^2 + y^2 = a^2 + b^2$. For hyperbola: $x^2 + y^2 = a^2 - b^2$ (exists only if $a > b$).
Worked Example
For the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$, find the equation of the director circle.
Director circle: $x^2 + y^2 = a^2 + b^2 = 25 + 16 = \boldsymbol{41}$
Any point on this circle has the property that the two tangent lines drawn from it to the ellipse are perpendicular.
Tip: Coordinate geometry is one of the most scoring topics in JEE Mains. Memorize the standard forms and tangent conditions for all three conics. Most problems reduce to plugging into these formulas once you identify the conic type.
Coordinate Geometry Formulas For JEE 2026: Conclusion
Coordinate Geometry is one of the most scoring sections in JEE Mathematics, as many questions are directly based on standard formulas and concepts. A strong understanding of lines, circles, and conic sections helps in solving problems quickly and accurately.
Regular practice along with consistent revision of formulas improves problem-solving speed and reduces mistakes. Focusing on key concepts like slopes, tangents, and properties of conics can significantly boost overall performance in the exam.
