Formula Quick-Reference

General Conic Properties

  • Locus Definition: SP=e⋅PMSP = e \cdot PM

  • Eccentricity (e): Circle (e=0e=0), Parabola (e=1e=1), Ellipse (0<e<10<e<1), Hyperbola (e>1e>1)

The Circle 🔵

  • Standard Equation: (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2

  • General Equation: x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0

  • Center: (−g,−f)(-g, -f)

  • Radius: r=g2+f2−cr = \sqrt{g^2+f^2-c}

  • Parametric Form: (h+rcos⁡θ,k+rsin⁡θ)(h+r\cos\theta, k+r\sin\theta)

The Parabola

  • Standard Equation: y2=4axy^2 = 4ax

  • Focus: (a,0)(a, 0)

  • Directrix: x=−ax = -a

  • Latus Rectum Length: 4a4a

  • Parametric Form: (at2,2at)(at^2, 2at)

The Ellipse

  • Standard Equation: x2a2+y2b2=1(a>b)\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a>b)

  • Key Relationship: c2=a2−b2c^2=a^2-b^2 and b2=a2(1−e2)b^2 = a^2(1-e^2)

  • Foci: (±c,0)(\pm c, 0) or (±ae,0)(\pm ae, 0)

  • Sum of Focal Distances: PF1+PF2=2aPF_1 + PF_2 = 2a

  • Latus Rectum Length: 2b2/a2b^2/a

  • Parametric Form: (acos⁡θ,bsin⁡θ)(a\cos\theta, b\sin\theta)

The Hyperbola

  • Standard Equation: x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

  • Key Relationship: c2=a2+b2c^2=a^2+b^2 and b2=a2(e2−1)b^2 = a^2(e^2-1)

  • Foci: (±c,0)(\pm c, 0) or (±ae,0)(\pm ae, 0)

  • Asymptotes: y=±baxy = \pm \frac{b}{a}x

  • Difference of Focal Distances: ∣PF1−PF2∣=2a|PF_1 - PF_2| = 2a

  • Latus Rectum Length: 2b2/a2b^2/a

  • Parametric Form: (asec⁡θ,btan⁡θ)(a\sec\theta, b\tan\theta)

🎯 JEE Exam Tips & Tricks

1. Know Your Conic's Anatomy

You must have the standard properties of each conic memorized. For any standard equation, you should be able to instantly state its vertices, foci, directrix equation, and latus rectum length.

Trick: Create a one-page formula sheet with a diagram for each of the four standard parabolas, two ellipses, and two hyperbolas. Visual association is key to quick recall during an exam.

2. Handle Shifted Conics with Ease

Many problems involve conics not centered at the origin. The key is to complete the square to get the equation into the form (x−h)2a2±(y−k)2b2=1\frac{(x-h)^2}{a^2} \pm \frac{(y-k)^2}{b^2} = 1 or (y−k)2=4a(x−h)(y-k)^2 = 4a(x-h).

Trick: To find parameters like the focus, think of the shifted conic as a standard one in a new coordinate system where X=(x−h)X = (x-h) and Y=(y−k)Y = (y-k). Find the parameter in the XYXY-plane (e.g., focus at (c,0)(c,0)) and then convert back to the original system: x−h=c  ⟹  x=h+cx-h=c \implies x=h+c.

3. Memorize Tangent Conditions

Questions on tangents are high-yield. Memorizing the slope-form conditions for tangency is a huge time-saver. The line y=mx+cy=mx+c is a tangent if:

  • Parabola (y2=4axy^2=4ax): c=a/mc = a/m

  • Ellipse (x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1): c2=a2m2+b2c^2 = a^2m^2+b^2

  • Hyperbola (x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1): c2=a2m2−b2c^2 = a^2m^2-b^2

4. Use Parametric Coordinates for Locus Problems

If a problem asks for the locus of a point related to a conic, using parametric coordinates is often the best approach. It turns a complex geometry problem into a simpler algebra problem of eliminating the parameter ('t' or 'θ').

5. Recognize Key Locus Properties

Knowing standard locus theorems can solve a problem in seconds. Two of the most important are:

  • Director Circle: The locus of the intersection of perpendicular tangents.

  • Ellipse: x2+y2=a2+b2x^2+y^2 = a^2+b^2

  • Hyperbola: x2+y2=a2−b2x^2+y^2 = a^2-b^2

  • Parabola: Its directrix, x=−ax=-a.

  • Auxiliary Circle: The locus of the foot of the perpendicular from a focus to any tangent. For both the ellipse and hyperbola, its equation is x2+y2=a2x^2+y^2=a^2.