Coordinate Geometry provides algebraic tools to study geometric figures. This topic is heavily tested in JEE.
Overview
graph TD
A[Coordinate Geometry] --> B[Basics]
A --> C[Straight Lines]
A --> D[Circles]
A --> E[Conic Sections]
E --> E1[Parabola]
E --> E2[Ellipse]
E --> E3[Hyperbola]Basic Concepts
Distance Formula
$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$Section Formula
Point dividing line joining $(x_1, y_1)$ and $(x_2, y_2)$ in ratio $m:n$:
Internal division:
$$\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$$External division:
$$\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}\right)$$Midpoint
$$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$Centroid of Triangle
$$G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$Straight Lines
Slope
$$m = \tan\theta = \frac{y_2-y_1}{x_2-x_1}$$Forms of Line Equation
| Form | Equation |
|---|---|
| Slope-intercept | $y = mx + c$ |
| Point-slope | $y - y_1 = m(x - x_1)$ |
| Two-point | $\frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}$ |
| Intercept | $\frac{x}{a} + \frac{y}{b} = 1$ |
| Normal | $x\cos\alpha + y\sin\alpha = p$ |
| General | $ax + by + c = 0$ |
Angle Between Two Lines
$$\boxed{\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|}$$Parallel lines: $m_1 = m_2$ Perpendicular lines: $m_1 \cdot m_2 = -1$
Distance of Point from Line
Distance of $(x_1, y_1)$ from $ax + by + c = 0$:
$$\boxed{d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}}$$Distance Between Parallel Lines
For $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$:
$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$For foot of perpendicular from $(x_1, y_1)$ to $ax + by + c = 0$:
$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$$Circles
Standard Form
$$\boxed{(x-h)^2 + (y-k)^2 = r^2}$$Center: $(h, k)$, Radius: $r$
General Form
$$x^2 + y^2 + 2gx + 2fy + c = 0$$Center: $(-g, -f)$, Radius: $\sqrt{g^2 + f^2 - c}$
Equation from Diameter Endpoints
If $(x_1, y_1)$ and $(x_2, y_2)$ are endpoints:
$$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$$Position of Point w.r.t. Circle
For point $(x_1, y_1)$ and circle $S = x^2 + y^2 + 2gx + 2fy + c$:
- $S_1 < 0$: Inside
- $S_1 = 0$: On circle
- $S_1 > 0$: Outside
Tangent to Circle
At point $(x_1, y_1)$ on circle $x^2 + y^2 = a^2$:
$$xx_1 + yy_1 = a^2$$Condition for tangency: Line $y = mx + c$ is tangent to $x^2 + y^2 = a^2$ if:
$$c^2 = a^2(1 + m^2)$$Parabola
Standard Form: $y^2 = 4ax$
| Element | Value |
|---|---|
| Vertex | $(0, 0)$ |
| Focus | $(a, 0)$ |
| Directrix | $x = -a$ |
| Latus rectum | $4a$ |
| Eccentricity | $e = 1$ |
Other Forms
| Equation | Opens |
|---|---|
| $y^2 = 4ax$ | Right |
| $y^2 = -4ax$ | Left |
| $x^2 = 4ay$ | Upward |
| $x^2 = -4ay$ | Downward |
Tangent
At $(x_1, y_1)$: $yy_1 = 2a(x + x_1)$
For $y = mx + c$ to be tangent: $c = \frac{a}{m}$
Ellipse
Standard Form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (a > b)
| Element | Value |
|---|---|
| Center | $(0, 0)$ |
| Foci | $(\pm c, 0)$ where $c = ae$ |
| Vertices | $(\pm a, 0)$ |
| Eccentricity | $e = \sqrt{1 - \frac{b^2}{a^2}}$ |
| Latus rectum | $\frac{2b^2}{a}$ |
| Directrix | $x = \pm\frac{a}{e}$ |
Key Relation
$$\boxed{b^2 = a^2(1 - e^2)}$$or
$$c^2 = a^2 - b^2$$Tangent
At $(x_1, y_1)$: $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$
Condition for tangency: $c^2 = a^2m^2 + b^2$
Hyperbola
Standard Form: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
| Element | Value |
|---|---|
| Center | $(0, 0)$ |
| Foci | $(\pm c, 0)$ where $c = ae$ |
| Vertices | $(\pm a, 0)$ |
| Eccentricity | $e = \sqrt{1 + \frac{b^2}{a^2}}$ |
| Latus rectum | $\frac{2b^2}{a}$ |
| Asymptotes | $y = \pm\frac{b}{a}x$ |
Key Relation
$$\boxed{b^2 = a^2(e^2 - 1)}$$or
$$c^2 = a^2 + b^2$$Rectangular Hyperbola
When $a = b$: $x^2 - y^2 = a^2$
- Eccentricity: $e = \sqrt{2}$
- Asymptotes: $y = \pm x$
Conjugate Hyperbola
$$\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$$Eccentricities of hyperbola and conjugate: $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$
Practice Problems
Find the equation of line passing through $(2, 3)$ perpendicular to $3x + 4y = 12$.
Find the radius of circle passing through $(3, 4)$ with center at $(1, 2)$.
Find the eccentricity of ellipse $4x^2 + 9y^2 = 36$.
Find the equation of tangent to parabola $y^2 = 8x$ parallel to $x - y + 3 = 0$.