Conic Sections Standard Form

A conic (section) is the locus of a point moving in a plane such that its distance from a fixed point (focus) is in a constant ratio to its perpendicular distance from a fixed line (i.e. directrix). This constant ratio is called eccentricity of the conic.

The eccentricity of a circle is zero. It shows how “un-circular” a curve is. Higher the eccentricity, lower curved it is.

Conic Sections

 

Terminology:

Axis of conic: Line passing through focus, perpendicular to the directrix.

Vertex: Point of the intersection of conic and axis.

Chord: Line segment joining any 2 points on the conic.

Double ordinate: Chord perpendicular to the axis

Latus Rectum: Double ordinate passing through focus.

Standard Parabola

Standard Equation Directrix Focus Length of Latus rectum Vertex
y2 = 4ax x = – a S : (a, 0) 4a (0, 0)
y2 = – 4ax x = a (- a, 0) 4a (0, 0)
x2 = 4ay y = – a (0, +a) 4a (0, 0)
x2 = – 4ay y = a (0, -a) 4a (0, 0)

 

Important results of a Parabola

1. 4 x distance between vertex and focus = Latus rectum = 4a.

2. 2 x Distance between directrix and focus = Latus rectum = 2(2a).

3. Point of intersection of Axis and directrix and the focus is bisected by the vertex.

4. Focus is the mid point of the Latus rectum.

5. (Distance of any point on parabola from axis)2 = (LR) (Distance of same point from tangent at vertex)

Focal Distance of a point

Focal Distance of a Parabola from a point

PS = PM = a + x1 = a + at2

PS=(x1a)2+(y12)PS=\sqrt{{{({{x}_{1}}-a)}^{2}}+({{y}_{1}}^{2})} =x12+a22ax1+(4ax1)=\sqrt{{{x}_{1}}^{2}+{{a}^{2}}-2a{{x}_{1}}+(4a{{x}_{1}})} =(x1+a)2=\sqrt{{{\left( {{x}_{1}}+a \right)}^{2}}} =x1+a=\left| {{x}_{1}}+a \right| =a+at2=\left| a+a{{t}^{2}} \right|

 

Equation of chord for y2 = 4ax

Slope PQm=2a(t2t1)a(t22t12)PQ\Rightarrow m=\frac{2a({{t}_{2}}-{{t}_{1}})}{a(t_{2}^{2}-t_{1}^{2})}

= 2t1+t2\frac{2}{{{t}_{1}}+{{t}_{2}}}

 

y2at1=2(t1+t2)(xat12)y-2a{{t}_{1}}=\frac{2}{({{t}_{1}}+{{t}_{2}})}\left( x-at_{1}^{2} \right) y(t1+t2)=2x+2at1t2\Rightarrow y({{t}_{1}}+{{t}_{2}})=2x+2a{{t}_{1}}{{t}_{2}}

 

Focal chord: Goes through (a, 0)

O(t1+t2)=2a+2at1t2\Rightarrow O({{t}_{1}}+{{t}_{2}})=2a+2a{{t}_{1}}{{t}_{2}} t1t2=1\Rightarrow {{t}_{1}}{{t}_{2}}=-1

Or

Productofordinates=2at1×2at2=4a2t1t2=4a2\text{Product}\,\text{of}\,\text{ordinates}=2a{{t}_{1}}\times 2a{{t}_{2}}=4{{a}^{2}}t_{1}^{{}}{{t}_{2}}=-4{{a}^{2}}.

 

Length of focal chord:

a(t2t1)2P(ab2,2ab)\frac{a{{({{t}_{2}}-{{t}_{1}})}^{2}}}{P(a{{b}^{2}},2ab)}

 

PQ=a2(t12t11)2+4a2(t1t2)2PQ=\sqrt{{{a}^{2}}\left( t_{1}^{2}-t_{1}^{1} \right)2+4{{a}^{2}}{{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}}

 

=at1t2(t2+t1)2+4=a\left| {{t}_{1}}-{{t}_{2}} \right|\sqrt{{{\left( {{t}_{2}}+{{t}_{1}} \right)}^{2}}+4}

 

=at1t2t12+t22+2t1t2+4(t1t2)(since,t1t2=1)=a\left| {{t}_{1}}-{{t}_{2}} \right|\sqrt{t_{1}^{2}+t{}_{2}^{2}+2t_1t_2+4(-{{t}_{1}}{{t}_{2}})}\,\,(since,\,{{t}_{1}}{{t}_{2}}=-1)

 

=at1t2t1t2=a\left| {{t}_{1}}-{{t}_{2}} \right|\left| {{t}_{1}}-t{}_{2} \right| = a (t2 – t1)2

PQ = PS + SQ

=at12+a+at22+a=at_{1}^{2}+a+at_{2}^{2}+a =2a+a(t12+(1t1)2)=2a+a\left( t_{1}^{2}+{{\left( \frac{-1}{{{t}_{1}}} \right)}^{2}} \right) =a(t12+1t12+2)=a(t1+1t1)21=a\left( t_{1}^{2}+\frac{1}{t_{1}^{2}}+2 \right)=a{{\left( {{t}_{1}}+\frac{1}{{{t}_{1}}} \right)}^{2}}\,\,\,\,\,\,\,\,……1

 

tanα=2t1+t2=2t11t1=2t1t121\tan \alpha =\frac{2}{{{t}_{1}}+{{t}_{2}}}=\frac{2}{{{t}_{1}}-\frac{1}{{{t}_{1}}}}=\frac{2t1}{t_{1}^{2}-1}

 

cot2α=cosec2α1=(t121)24t12{{\cot }^{2}}\alpha =\cos e{{c}^{2}}\alpha -1=\frac{{{(t_{1}^{2}-1)}^{2}}}{4t_{1}^{2}} cosecα=(t111)2+4t124t12=12t14+2t12+1t12\cos ec\alpha =\sqrt{\frac{{{\left( t_{1}^{1}-1 \right)}^{2}}+4t_{1}^{2}}{4t_{1}^{2}}}=\frac{1}{2}\sqrt{\frac{t_{1}^{4}+2t_{1}^{2}+1}{t_{1}^{2}}} =12(t1+1t1)=\frac{1}{2}\left| \left( {{t}_{1}}+\frac{1}{{{t}_{1}}} \right) \right|

From (1),

PQ = a(2cosecα)2 = 4acosec2α

Ellipse

It is a locus of a point which moves such that the ratio of its distance from a fixed point (focus) to its distance from a fixed line (directrix) is always constant and less than 1, i.e o < e < 1.

Ellipse Introduction

PSPM=e<1\frac{PS}{PM}=e<1

 

Ellipse with Horizontal major axis

x2a2+y2b2=1;b<a\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1;b<a

Focus : There are 2 focii; (ae, 0) and (-ae, 0)

Directrix: These foci have corresponding directrices as x++ae&x=aerespectivelyx+\frac{+a}{e}\And x=\frac{-a}{e}respectively e2=1b2a2{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}

Axes:

xx’ : Major Axis: Length: 2a

yy’ : Minor axis; length : 2b

Vertox : (a, 0) & (-a, 0)

Centre : (0, 0)

Latus rectum: y=±b2/Ry=\pm {}^{b{}^{2}}/{}_{R} (solvex=±aewithx2a2+y2b2=1)\left( solve\,x=\pm \,ae\,with\,\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 \right)

Length of Latus rectum: y=2b2/ay=2b{}^{2}/{}_{a}

 

Ellipse with vertical major axis

x2a2+y2b2=1;b>a\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1;b>a e2=1a2b2\Rightarrow {{e}^{2}}=1-\frac{{{a}^{2}}}{{{b}^{2}}}

Length of major axis : 2b

Length of minor axis : 2a

Focii : (0, be) & (0, -be)

Directrices: x = b/e & x = -b/e

Latus rectum: y = + be

Length of Latus rectum = 2a2b\frac{2{{a}^{2}}}{b}

 

Important results

1. e=1(semi minor axis)2(semi major axis)2e=\sqrt{1-\frac{{{(semi\ minor\ axis)}^{2}}}{{{(semi\ major\ axis)}^{2}}}}

 

2. Length of Latus Rectum=2(semi minor axis)2)(semi minor axisLength\ of\ Latus\ Rectum=\frac{2(semi\ minor\ axis)^2)}{(semi\ minor\ axis}

 

3. Distance between 2 directices: Majoraxiseccentricity\frac{Major\,axis}{eccentricity}

4. Distance between 2 focii: (major axis) × eccentricity

5. Distance between focus and directrix: aeaeorbebe\frac{a}{e}-ae\,\,or\,\,\frac{b}{e}-be

Hyperbola

If is the locus of a point which moves such that the ratio of its distance from a fixed point (focus) to its distance from a fixed line (directrix) is always constant and greater than 1.

PSPM=e>1\frac{PS}{PM}=e>1

Hyperbola Introduction

Standard Hyperbola:

x2a2y2b2=1\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1

Focus: There are 2 focii (ae, 0) and (-ae, 0)

Directrix : The foci has corresponding directrices as x = +a/e and x = −a/e respectively.

e2=1+b2a2;b2=a2(e21){{e}^{2}}=1+\frac{{{b}^{2}}}{{{a}^{2}}};{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)

Axis: xx’ : Transverse axis ; Length : 2a

yy’ : Conjugate axis ; Length : 2b (Hypothetical)

Vertex (0, 0) and (-a, 0)

Centre (0, 0)

Latus rectum: x = +ae

Length of latus rectum = 2b2a\frac{2{{b}^{2}}}{a}

 

Position of point at hyperbola x2a2y2b2=1\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1

Let S=x2a2y2b21S=\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}-1

and S1=x12a2y12b21{{S}_{1}}=\frac{{{x}_{1}}^{2}}{{{a}^{2}}}-\frac{{{y}_{1}}^{2}}{{{b}^{2}}}-1

If S1 > 0, point C lies inside the hyperbola

S1 = 0 point B lies on the hyperbola

S1 < 0 point A lies outside the hyperbola.

RECOMMENDED VIDEOS

Parabola

Ellipse and Hyperbola

Solved Examples

Example 1: Find equation of a conic whose focus is at (1, 0) and directrix is 2x + 5y + 1 = 0. Also, e=1/2e=1/\sqrt{2}

Solution: 12=(h1)2+k212h+5k+1129\frac{1}{\sqrt{2}}=\frac{\sqrt{{{(h-1)}^{2}}+{{k}^{2}}}}{\frac{12h+5k+11}{\sqrt{29}}} 12x+5y+1158=(x1)2+y2\frac{12x+5y+11}{\sqrt{58}}=\sqrt{{{(x-1)}^{2}}+{{y}^{2}}}

Squaring both sides, we get

4x2 + 25y2 + 1 + 20xy + 10y + 4x

Which is the required equation.

Example 2: If extreme points of LR are (11/2, 6) and (13/2, 4). Find the equation of the parabola.

Solution:

Mid point of LR : focus : (6, 5)

Now, 4a = 2 or a = ½

The equation of parabolas are:

(y – 5)2 = 2(x – 5.5) and (y – 5)2 = – 2 (x – 6.5)