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Coordinate Geometry JEE Notes: Important Concepts
The chapter deals with problems on a two-dimensional Cartesian plane, but questions frequently combine algebraic manipulation, graphical insight and geometric reasoning. JEE typically checks:
- Distance & Division: quick ratio reasoning, midpoint and section formula.
- Slope and Angle: finding or comparing inclinations, perpendicular and parallel conditions.
- Locus: converting a geometric sentence into an equation and vice-versa.
- Family of Curves: expressing lines/circles/conics passing through points or satisfying constraints.
- Transformations: shifting or rotating axes, translating curves.
Master these fundamentals first, then move to curve-specific formulae. Almost every recent paper reserved 3-4 percent of the total marks for a conceptual but computation-light coordinate geometry question.
Straight Line and Pair of Lines
Key Results
| Concept | Result / Condition |
|---|---|
| Distance between $$(x_1,y_1)$$ and $$(x_2,y_2)$$ | $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ |
| Section formula (internal) dividing $$P_1P_2$$ in ratio $$m:n$$ | $$\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)$$ |
| Slope of line joining $$(x_1,y_1),(x_2,y_2)$$ | $$m=\frac{y_2-y_1}{x_2-x_1}$$ if $$x_2\neq x_1$$ |
| Angle $$\theta$$ between lines $$m_1,m_2$$ | $$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$$ |
| Normal form | $$x\cos\alpha+y\sin\alpha=p$$ |
| Perpendicular distance from $$(x_1,y_1)$$ to $$ax+by+c=0$$ | $$\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$$ |
| Pair of lines from second-degree homogeneous $$ax^2+2hxy+by^2=0$$ | Slopes are roots of $$bm^2+2hm+a=0$$ |
Worked Example 1 (Slope, Angle & Distance)
Find the angle between the lines joining the origin to the points $$A(2,3)$$ and $$B(-4,1)$$. Also find the length $$AB$$.
- Slopes: $$m_1=\frac{3}{2}$$, $$m_2=\frac{1}{-4}=-\frac14$$.
- Angle: $$\tan\theta=\left|\frac{-\frac14-\frac32}{1+(-\frac14)(\frac32)}\right|=\left|\frac{-\frac{7}{4}}{1-\frac38}\right|=\frac{7/4}{5/8}= \frac{7}{4}\cdot\frac{8}{5}=\frac{14}{5}$$ so $$\theta=\tan^{-1}\left(\frac{14}{5}\right)$$.
- Distance $$AB=\sqrt{(2+4)^2+(3-1)^2}= \sqrt{6^2+2^2}= \sqrt{36+4}= \sqrt{40}=2\sqrt{10}$$.
Answer: $$\theta=\tan^{-1}(14/5)$$, $$AB=2\sqrt{10}$$.
Worked Example 2 (Family of Lines)
Find the equation of the straight line through the intersection of $$3x+4y-5=0$$ and $$x-2y+1=0$$ that makes an angle $$45^\circ$$ with the x-axis.
- Intersection point: $$\begin{cases}3x+4y=5\\x-2y=-1\end{cases} \Rightarrow x= \frac{8}{11},\; y=\frac{19}{11}$$.
- Slope required: $$m=\tan45^\circ=1$$.
- Equation: $$y-\frac{19}{11}=1\left(x-\frac{8}{11}\right) \Rightarrow 11y-19=11x-8$$ or $$11x-11y+11=0$$.
Answer: $$11x-11y+11=0$$.
After finishing every sub-topic, try 10 past numerical problems. The curated list inside JEE Questions lets you filter exactly “Straight Line” so you do not waste time hunting.
Circle and System of Circles
Standard Equations & Properties
| Form | Equation | Main Parameter |
|---|---|---|
| Centre-radius | $$(x-h)^2+(y-k)^2=r^2$$ | Centre $$(h,k)$$, radius $$r$$ |
| General | $$x^2+y^2+2gx+2fy+c=0$$ | Centre $$(-g,-f)$$, $$r=\sqrt{g^2+f^2-c}$$ |
| Diameter AB | $$(xx_1+yy_1)-(x+y)(x_1+y_1)+x_1y_1+xy=0$$ | Endpoints $$A(x_1,y_1),\;B(x_2,y_2)$$ |
| Family through two points $$S=0, S'=0$$ | $$S+\lambda S'=0$$ | Real $$\lambda$$ gives required circle |
| Radical axis of two circles | Subtract general equations $$S_1-S_2=0$$ | Always a straight line |
| Length of tangent from $$P(x_1,y_1)$$ | $$S_{1}=x_1^2+y_1^2+2gx_1+2fy_1+c$$ |
For any point $$P(x_1,y_1)$$, if $$S_1=0$$ the point lies on the circle, $$S_1<0$$ is inside and $$S_1>0$$ is outside.
Worked Example 3 (Tangent & Radical Axis)
Find the equation of the radical axis of circles $$x^2+y^2-4x+6y-5=0$$ and $$x^2+y^2+2x-4y-23=0$$.
Subtract: $$(x^2+y^2-4x+6y-5)-(x^2+y^2+2x-4y-23)=0$$.
Simplify to $$-6x+10y+18=0$$ or $$3x-5y-9=0$$.
Answer: $$3x-5y-9=0$$.
Worked Example 4 (Circle through Three Points)
Find the circle passing through $$A(2,1),\;B(-1,3),\;C(4,5)$$ and determine whether the origin lies inside it.
- Use determinant form:
$$ \begin{vmatrix} x^2+y^2 & x & y & 1\\ 2^2+1^2 & 2 & 1 & 1\\ (-1)^2+3^2 & -1 & 3 & 1\\ 4^2+5^2 & 4 & 5 & 1 \end{vmatrix}=0 $$ Expanding, we finally get $$x^2+y^2-6x-4y-3=0$$. - Centre $$C_0(3,2)$$, radius $$r=\sqrt{9+4+3}= \sqrt{16}=4$$.
- Origin test: $$S_1=0^2+0^2-6(0)-4(0)-3=-3<0$$ so the origin is inside.
Answer: Circle $$x^2+y^2-6x-4y-3=0$$, origin is inside.
Complex multi-constraint questions of this sort often appear in Paper 2 of Advanced. Cross-check your approach with full solutions in JEE Advanced Previous Papers whenever you get stuck.
Conic Sections: Parabola, Ellipse, Hyperbola
Standard Forms
| Curve | Standard Equation (centre at origin) | Focus | Directrix |
|---|---|---|---|
| Parabola (opening right) | $$y^2=4ax$$ | $$(a,0)$$ | $$x=-a$$ |
| Ellipse (major axis x) | $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\;a>b$$ | $$(\pm c,0),\;c^2=a^2-b^2$$ | $$x=\pm\frac{a^2}{c}$$ |
| Hyperbola (transverse x) | $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ | $$(\pm c,0),\;c^2=a^2+b^2$$ | $$x=\pm\frac{a^2}{c}$$ |
Auxiliary Points and Parameters
- Parabola parameter $$t$$: point $$P(at^2,2at)$$ on $$y^2=4ax$$.
- Ellipse parameter $$\theta$$: point $$P(a\cos\theta,b\sin\theta)$$.
- Hyperbola parameter $$\theta$$: point $$P(a\sec\theta,b\tan\theta)$$.
- Eccentricity: $$e=1$$ for parabola, $$e<1$$ ellipse, $$e>1$$ hyperbola.
Tangent, Normal & Chord Equations
| Curve | Tangent at $$P$$ | Normal at $$P$$ |
|---|---|---|
| Parabola $$y^2=4ax$$ with parameter $$t$$ | $$ty=x+at^2$$ | $$y+tx=2at+at^3$$ |
| Ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ | $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$$ | $$a^2y_1y-b^2x_1x=a^2b^2(x_1y_1-r^2)$$ |
| Hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ | $$\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1$$ | Similar swap of sign in ellipse normal |
Directrix–Focus Property
For all three conics, the ratio (distance of point from focus)/(distance from directrix) equals the eccentricity $$e$$. This is the quickest path to write a locus when JEE disguises the curve in words.
Worked Example 5 (Parabola Tangent)
Find the points on $$y^2=8x$$ where the tangents are inclined at $$45^\circ$$ to the x-axis.
- Slope required: $$m=\tan45^\circ=1$$.
- Tangent form: $$y=mx+\frac{2a}{m}$$, here $$a=2$$, so $$y=x+\frac{4}{1}=x+4$$.
- Intersect with parabola: Substitute $$y=x+4$$ into $$y^2=8x$$: $$(x+4)^2=8x \Rightarrow x^2+8x+16=8x \Rightarrow x^2+16=0$$.
No real $$x$$ satisfies the equation, hence no real tangent of slope 1 exists. The question’s answer is no real point. This trick reappears often; always check solvability.
Worked Example 6 (Ellipse Area of Sector)
For the ellipse $$\frac{x^2}{16}+\frac{y^2}{9}=1$$, find the area swept from the major axis to the line joining the centre to the point corresponding to $$\theta=\frac{\pi}{3}$$.
- Parametric point $$P(4\cos\theta,3\sin\theta)\Rightarrow P\left(2, \frac{3\sqrt{3}}{2}\right)$$.
- Area of sector of ellipse: $$A=\frac{ab}{2}(\theta)$$ by calculus (JEE favourite derivation).
- $$A=\frac{4\times3}{2}\left(\frac{\pi}{3}\right)=6\cdot\frac{\pi}{3}=2\pi$$.
Answer: $$2\pi$$ square units.
Keep a personal sheet of all such small but exotic formulas. You can cross-verify any you miss with the Maths PDF on JEE Formula Sheets.
Worked Example 7 (Rectangular Hyperbola Transformation)
The parabola $$y^2=4ax$$ is translated by $$(-a,0)$$ and rotated by $$45^\circ$$ anticlockwise. Show that the new curve is a rectangular hyperbola and find its equation.
- Translate: replace $$x\to x+a$$ to get $$y^2=4a(x+a)$$.
- Rotate: $$x=X\cos45^\circ-Y\sin45^\circ,\;y=X\sin45^\circ+Y\cos45^\circ$$.
- Substitute and simplify to $$XY=a^2$$.
Since the product of rotated coordinates is a constant, the curve is a rectangular hyperbola.
Important Formulas and Results at a Glance
| Theme | Ready-to-use Formula | Best-in-Exam Use |
|---|---|---|
| Centroid of triangle $$ABC$$ | $$(x_G,y_G)=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right)$$ | Interior ratio 2:1 with any median |
| Incentre | $$\left(\frac{ax_A+bx_B+cx_C}{a+b+c},\frac{ay_A+by_B+cy_C}{a+b+c}\right)$$ | Quick radius $$r=\Delta/s$$ |
| Area of polygon by coordinates | $$\frac12\left|\sum x_iy_{i+1}-x_{i+1}y_i\right|$$ | Works wonders in objective questions |
| Director circle of ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ | $$x^2+y^2=a^2+b^2$$ | Perpendicular tangents locus |
| Chord with midpoint $$M(x_1,y_1)$$ in circle $$S=0$$ | $$T=0$$ method: $$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$$ | Eliminates parameters quickly |
| Polar equation of line w.r.t. circle | $$T=0$$ interchange trick | Fastest for combined geometry-calculus MCQs |
Must-memorise triplet: $$y^2=4ax,\;\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\;\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$. Tangent, normal, length of latus rectum and focal distance from any point can be written in under 5 seconds when you know them cold.
Once you finish revising the above list, solve at least one full chapter set from JEE Mains Previous Papers. You will notice that 70 percent of the answers need nothing beyond these boxed equations.
JEE Important Points, Common Mistakes and Quick Revision
Exam-Wise Weightage
| Sub-topic | Typical JEE Mains marks | Typical JEE Adv marks |
|---|---|---|
| Straight line (slope, distance, family) | 4 | 6 |
| Circle & system | 4 | 8 |
| Parabola | 4 | 6 |
| Ellipse + Hyperbola combo | 0-4 | 8 |
Common Student Slips
- Forgetting that the angle formula $$\tan\theta$$ fails when lines are perpendicular but $$1+m_1m_2=-1$$ numerator/denominator sign can be mishandled.
- Mixing up $$a$$ and $$b$$ between ellipse and hyperbola; always write $$c^2=a^2\pm b^2$$ at the top of rough sheet before starting.
- Using the centre-radius form for a shifted circle without expanding the square properly, leading to sign errors.
- Not squaring both sides when applying distance ratio definitions for locus questions, thus dropping extraneous solutions.
- Ignoring domain restrictions of parameters: $$\theta$$ for ellipse is not the same as $$t$$ for parabola.
30-Second Checklist Before Submitting Paper
- Have you copied every constant correctly after expanding? A lost “-” changes the curve completely.
- Verify units: if the question asks for area, your answer cannot contain radicals inside distance units.
- Circle radius squared must be positive; if you get $$r^2\lt0$$ stop—you made an algebra slip.
- When locus is asked, confirm the final equation is free from parameter $$t,\theta,\lambda$$.
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