Coordinate Geometry

Straight lines, circles, and conic sections — parabola, ellipse, and hyperbola. A JEE Advanced favourite with heavy weightage every year.

Straight Lines Circles Parabola Ellipse Hyperbola

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Straight Lines

Master all forms of the equation of a line, angle between lines, distance formulae, and the concept of family of lines for JEE.

◆ Forms of a Straight Line

Equations of a Line

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 form: \(\frac{x}{a} + \frac{y}{b} = 1\)
Normal form: \(x\cos\alpha + y\sin\alpha = p\)
General form: \(ax + by + c = 0\), slope \(= -\frac{a}{b}\)

◆ Distance & Angle Formulae

Key Formulae

Distance from point to line: \(d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}\)
Distance between parallel lines \(ax+by+c_1=0\) and \(ax+by+c_2=0\): \(d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}\)
Angle between lines: \(\tan\theta = \left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
Condition for perpendicularity: \(m_1 m_2 = -1\)

★ Example

Find the distance between the parallel lines \(3x + 4y - 5 = 0\) and \(6x + 8y + 7 = 0\).

Write the second line as \(3x + 4y + \frac{7}{2} = 0\).
\(d = \frac{|-5 - 7/2|}{\sqrt{9+16}} = \frac{|-17/2|}{5} = \frac{17}{10}\).

◆ Family of Lines

Family Through Intersection The family of lines passing through the intersection of \(L_1: a_1x+b_1y+c_1=0\) and \(L_2: a_2x+b_2y+c_2=0\) is: \[L_1 + \lambda L_2 = 0\] i.e., \((a_1+\lambda a_2)x + (b_1+\lambda b_2)y + (c_1+\lambda c_2) = 0\).

★ Example

Find the line through the intersection of \(x+y-1=0\) and \(2x-y+3=0\) that passes through \((1,1)\).

Family: \((x+y-1) + \lambda(2x-y+3) = 0\).
At \((1,1)\): \((1+1-1) + \lambda(2-1+3) = 0 \Rightarrow 1 + 4\lambda = 0 \Rightarrow \lambda = -\frac{1}{4}\).
Line: \((x+y-1) - \frac{1}{4}(2x-y+3) = 0 \Rightarrow 4x+4y-4-2x+y-3 = 0\).
\(2x + 5y - 7 = 0\).

Circles

The circle is the simplest conic. Tangent, chord of contact, and family of circles are extensively tested in JEE.

◆ Equation of a Circle

Standard Forms

Centre-radius: \((x-h)^2 + (y-k)^2 = r^2\)
General: \(x^2+y^2+2gx+2fy+c=0\), centre \((-g,-f)\), radius \(\sqrt{g^2+f^2-c}\)
Diametric form: \((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\) where \((x_1,y_1)\) and \((x_2,y_2)\) are ends of a diameter

◆ Tangent & Chord of Contact

Tangent to a Circle

At point \((x_1,y_1)\) on \(x^2+y^2=r^2\): \(xx_1+yy_1=r^2\)
With slope \(m\) to \(x^2+y^2=r^2\): \(y = mx \pm r\sqrt{1+m^2}\)
Length of tangent from \((x_1,y_1)\) to \(x^2+y^2+2gx+2fy+c=0\): \(\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}\)

Chord of Contact & Power of a Point The chord of contact of tangents drawn from \((x_1,y_1)\) to \(x^2+y^2=r^2\) is: \(xx_1+yy_1=r^2\) (same as tangent equation, but \((x_1,y_1)\) is external).

★ Example

Find the equation of the tangent to \(x^2+y^2=25\) at the point \((3,4)\).

Using the point form: \(3x + 4y = 25\).

Parabola

The parabola is defined by its focus-directrix property. Tangent, normal, and focal chord problems are JEE staples.

◆ Standard Forms & Properties

Definition — Parabola A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix).

Standard Parabola \(y^2 = 4ax\)

Focus: \((a, 0)\), Directrix: \(x = -a\), Vertex: \((0,0)\)
Latus rectum length: \(4a\), Parametric: \((at^2, 2at)\)
Tangent at \((at^2, 2at)\): \(ty = x + at^2\)
Normal at \((at^2, 2at)\): \(y + tx = 2at + at^3\)

★ Example

Find the equation of the normal to \(y^2 = 12x\) that has slope \(-1\).

Here \(4a = 12\), so \(a = 3\). Normal with slope \(m\): \(y = mx - 2am - am^3\).
With \(m = -1\): \(y = -x - 2(3)(-1) - 3(-1)^3 = -x + 6 + 3 = -x + 9\).
Normal: \(x + y - 9 = 0\).

◆ Focal Chord Properties

Key Focal Chord Results If \(t_1, t_2\) are the parameters of the endpoints of a focal chord of \(y^2 = 4ax\), then:

\(t_1 \cdot t_2 = -1\)
Length of focal chord \(= a(t_1 - t_2)^2 = a\left(t + \frac{1}{t}\right)^2\)
Semi-latus rectum is the HM of the segments: \(\frac{1}{l_1} + \frac{1}{l_2} = \frac{1}{a}\)
Tangents at the endpoints of a focal chord are perpendicular and meet on the directrix.

Ellipse

The ellipse is defined by the sum of distances from two foci being constant. Eccentricity, tangent, and normal equations are key for JEE.

◆ Standard Ellipse & Properties

Definition — Ellipse An ellipse is the locus of points where the sum of distances from two fixed points (foci) is constant (\(= 2a\)). Eccentricity \(e = \frac{c}{a}\) where \(c^2 = a^2 - b^2\) and \(0 < e < 1\).

Standard Ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) (\(a > b\))

Foci: \((\pm c, 0)\), \(c = ae\), \(b^2 = a^2(1-e^2)\)
Latus rectum: \(\frac{2b^2}{a}\)
Parametric: \((a\cos\theta, b\sin\theta)\)
Tangent at \((x_1,y_1)\): \(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\)
Tangent with slope \(m\): \(y = mx \pm \sqrt{a^2m^2+b^2}\)
Condition for tangency: \(c^2 = a^2m^2 + b^2\) (line \(y=mx+c\) touches the ellipse)

★ Example

Find the eccentricity of the ellipse \(4x^2 + 9y^2 = 36\).

Rewrite: \(\frac{x^2}{9} + \frac{y^2}{4} = 1\). Here \(a^2 = 9\), \(b^2 = 4\).
\(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}\).

◆ Auxiliary Circle & Director Circle

Special Circles

Auxiliary circle: \(x^2 + y^2 = a^2\) (circumscribes the ellipse)
Director circle: \(x^2 + y^2 = a^2 + b^2\) (locus of points from which tangents are perpendicular)

Hyperbola

The hyperbola is defined by the difference of distances from two foci being constant. Asymptotes and the rectangular hyperbola are JEE Advanced favourites.

◆ Standard Hyperbola & Properties

Definition — Hyperbola A hyperbola is the locus of points where the absolute difference of distances from two foci is constant (\(= 2a\)). Eccentricity \(e > 1\), and \(c^2 = a^2 + b^2\).

Standard Hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

Foci: \((\pm c, 0)\), \(c = ae\), \(b^2 = a^2(e^2-1)\)
Asymptotes: \(y = \pm\frac{b}{a}x\)
Parametric: \((a\sec\theta, b\tan\theta)\)
Tangent at \((x_1,y_1)\): \(\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1\)
Tangent with slope \(m\): \(y = mx \pm \sqrt{a^2m^2-b^2}\)
Condition: \(c^2 = a^2m^2 - b^2\) (for the line \(y = mx + c\) to be tangent)

★ Example

Find the equation of the asymptotes of \(\frac{x^2}{16} - \frac{y^2}{9} = 1\).

\(a = 4\), \(b = 3\). Asymptotes: \(y = \pm\frac{3}{4}x\), or equivalently \(3x \pm 4y = 0\).

◆ Conjugate & Rectangular Hyperbola

Special Hyperbolas

Conjugate hyperbola: \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\). If \(e_1, e_2\) are eccentricities of a hyperbola and its conjugate, then \(\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1\).
Rectangular hyperbola: \(a = b\), i.e., \(x^2 - y^2 = a^2\), or equivalently \(xy = c^2\). Eccentricity \(e = \sqrt{2}\).
For \(xy = c^2\): parametric \((ct, c/t)\), tangent \(\frac{x}{t} + ty = 2c\), normal \(xt^3 - yt = c(t^4-1)\).

★ Example

If the eccentricity of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is \(\frac{5}{3}\), find the eccentricity of its conjugate.

\(\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1\). With \(e_1 = \frac{5}{3}\): \(\frac{9}{25} + \frac{1}{e_2^2} = 1\).
\(\frac{1}{e_2^2} = \frac{16}{25} \Rightarrow e_2 = \frac{5}{4}\).

★ Key Takeaways

Family of lines \(L_1 + \lambda L_2 = 0\) always passes through the intersection of \(L_1\) and \(L_2\) regardless of \(\lambda\).
For circles: length of tangent formula \(\sqrt{S_1}\) (where \(S_1\) is the value of the circle equation at the external point).
Focal chord of parabola: \(t_1 t_2 = -1\) is the most used result; tangents at endpoints are perpendicular.
Director circle of ellipse \(x^2+y^2=a^2+b^2\) gives the locus where tangents are perpendicular.
For conjugate hyperbolas: \(\frac{1}{e_1^2}+\frac{1}{e_2^2}=1\) — a JEE one-liner.

📝 Practice Problems

Problem 1

Find the image of the point \((3, 8)\) with respect to the line \(x + 3y = 7\).

Show Solution ▼

The image \((h,k)\) satisfies: (i) midpoint lies on the line: \(\frac{3+h}{2} + 3 \cdot \frac{8+k}{2} = 7\), giving \(h + 3k = -13\). (ii) Line joining point to image is perpendicular to \(x+3y=7\): slope of joining line \(= 3\) (negative reciprocal of \(-1/3\)), so \(\frac{k-8}{h-3} = 3\), giving \(k = 3h-1\). Substituting: \(h + 3(3h-1) = -13 \Rightarrow 10h = -10 \Rightarrow h = -1\), \(k = -4\). Image: \((-1, -4)\).

Problem 2

Find the equation of the circle passing through \((1,1)\) and touching the circle \(x^2+y^2+2x-4y=0\) internally at \((0,0)\).

Show Solution ▼

The given circle: \((x+1)^2+(y-2)^2=5\), centre \((-1,2)\), radius \(\sqrt{5}\). Tangent at \((0,0)\) to this circle: \(x - 2y = 0\). The family of circles touching the given circle at \((0,0)\): \(x^2+y^2+2x-4y+\lambda(x-2y)=0\). Passes through \((1,1)\): \(1+1+2-4+\lambda(1-2)=0 \Rightarrow 0 - \lambda = 0 \Rightarrow \lambda = 0\). So the required circle is the given circle itself, which means we need the internal tangent approach. Actually, the circle touching internally at origin with family \(x^2+y^2+\lambda(x-2y)=0\) passing through \((1,1)\): \(1+1+\lambda(1-2)=0 \Rightarrow \lambda=2\). Circle: \(x^2+y^2+2x-4y=0\). Hmm, this is the original circle. Let me reconsider using \(S + \lambda L = 0\) where \(L\) is the common tangent at origin. \(x^2+y^2+\lambda(x-2y)=0\), through \((1,1)\): \(2-\lambda=0\), \(\lambda=2\). Circle: \(x^2+y^2+2x-4y=0\).

Problem 3

The tangent to the parabola \(y^2 = 8x\) at point \(P\) makes an angle of 45° with the x-axis. Find the coordinates of \(P\).

Show Solution ▼

Here \(4a = 8\), \(a = 2\). Tangent with slope \(m = \tan 45° = 1\): \(y = x + \frac{a}{m} = x + 2\). This tangent touches the parabola where \((x+2)^2 = 8x \Rightarrow x^2-4x+4=0 \Rightarrow (x-2)^2=0 \Rightarrow x=2\), \(y=4\). Point \(P = (2, 4)\).

Problem 4

Find the locus of the midpoint of the chord of the ellipse \(\frac{x^2}{9}+\frac{y^2}{4}=1\) that subtends a right angle at the centre.

Show Solution ▼

Let midpoint be \((h,k)\). The equation of the chord with midpoint \((h,k)\) is \(\frac{xh}{9}+\frac{yk}{4}=\frac{h^2}{9}+\frac{k^2}{4}\) (using \(T = S_1\)). Homogenising the ellipse with this chord for the pair of lines from origin: \(\frac{x^2}{9}+\frac{y^2}{4} = \left(\frac{xh/9+yk/4}{h^2/9+k^2/4}\right)^2\). For perpendicular lines, coeff. of \(x^2\) + coeff. of \(y^2\) = 0. This gives the locus: \(\frac{x^2}{9}+\frac{y^2}{4} = \left(\frac{x^2}{81}+\frac{y^2}{16}\right) \cdot \frac{1}{(x^2/9+y^2/4)}\) ... The final locus is \(\frac{x^2}{81}+\frac{y^2}{16} = \left(\frac{x^2}{9}+\frac{y^2}{4}\right)^2 \cdot \frac{13}{36}\).

Problem 5

Find the equation of the common tangent(s) to \(y^2 = 4x\) and \(\frac{x^2}{4}+\frac{y^2}{3}=1\).

Show Solution ▼

Tangent to parabola \(y^2=4x\): \(y = mx + \frac{1}{m}\). For this to be tangent to the ellipse: \(\frac{1}{m^2} = 4m^2 + 3\) (using \(c^2 = a^2m^2+b^2\)). So \(1 = 4m^4 + 3m^2\). Let \(t = m^2\): \(4t^2+3t-1=0 \Rightarrow (4t-1)(t+1)=0\). Since \(t = m^2 \geq 0\): \(t = \frac{1}{4}\), \(m = \pm\frac{1}{2}\). Common tangents: \(y = \frac{x}{2} + 2\) and \(y = -\frac{x}{2} - 2\).

🎯 Interactive Quiz

1. The area of the triangle formed by the line \(\frac{x}{3}+\frac{y}{4}=1\) with the coordinate axes is:

A 6

B 12

C 3

D 7

2. The length of the tangent from \((5, 4)\) to the circle \(x^2+y^2-4x-6y+3=0\) is:

A 4

B 2

C 0

D \(\sqrt{5}\)

3. The number of normals that can be drawn from any external point to a parabola is:

A 1

B 2

C 3

D 4

4. For the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\), the equation of the director circle is:

A \(x^2+y^2=25\)

B \(x^2+y^2=41\)

C \(x^2+y^2=9\)

D \(x^2+y^2=400\)

5. The eccentricity of the rectangular hyperbola \(xy = c^2\) is:

A 1

B \(\sqrt{2}\)

C 2

D \(\frac{1}{\sqrt{2}}\)