A parabola is a Ushaped plane curve where any point is at an equal distance from a fixed point (known as the focus) and from a fixed straight line which is known as the directrix. Parabola is an integral part of conic section topic and all its concepts parabola areÂ covered here.
TABLE OF CONTENTS
 Definition
 Standard Equation
 Latus Rectum
 Parametric coordinates
 General Equations
 Tangent to a Parabola
 Normal to a Parabola
 Focal Chord Properties
 Focal Chord, Tangent and Normal Properties
 Forms
 Questions
Download this lesson as PDF:Parabola PDF
What is Parabola?
Section of a right circular cone by a plane parallel to a generator of the cone is a parabola. It is a locus of a point, which moves so that distance from a fixed point (focus) is equal to the distance from a fixed line (directrix)
 Fixed point is called focus
 Fixed line is called directrix
Standard Equation of Parabola
The simplest equation of a parabola is y^{2 }= x when the directrix is parallel to the yaxis. In general, if the directrix is parallel to the yaxis in the standard equation of a parabola is given as:
y^{2}Â = 4ax 
If the parabola is sideways i.e., the directrix is parallel to xaxis, the standard equation of a parabole becomes,
x^{2}Â = 4ay 
Apart from these two, the equation of a parabola can also beÂ y^{2}Â = 4ax andÂ x^{2}Â = 4ayÂ if the parabola is in the negative quadrants. Thus, the four equations of a parabola are given as:

 y^{2}Â = 4ax
 y^{2}Â = – 4ax
 x^{2}Â = 4ay
 x^{2}Â = – 4ay
Parabola Equation Derivation
In the above equation, “a” is the distance from the origin to the focus. Below is the derivation for the parabola equation. First, refer to the image given below.
From definition,
\(\frac{SP}{PM}=1\)
SP = PM
\(\sqrt{{{\left( xa \right)}^{2}}+{{y}^{2}}}=\left \frac{x+a}{1} \right\)
(x – a)^{2} + y^{2} = (x + a)^{2 }
\(y^{2}=4ax\) â‡’ Standard equation of Parabola.
Latus Rectum of Parabola
The latus rectum of a parabola is the chord that passes through the focus and is perpendicular to the axis of the parabola.
LSLâ€™ Latus Ractum
= \(2\left( \sqrt{4a.a} \right)\)
= 4a (length of latus Rectum)
Note: – Two parabola are said to be equal if their latus rectum are equal.
Parametric coordinates of Parabola
For a parabola, the equation isÂ y^{2} = 4ax. Now, to represent the coordinates of a point on the parabola, the easiest form will be = at^{2Â }and y = 2at as for any value of “t”, the coordinates (at^{2}, 2at) will always satisfy the parabola equation i.e. y^{2} = 4ax. So,
Any point on the parabola
y^{2} = 4ax (at^{2}, 2at)
where â€˜tâ€™ is a parameter.
Video Lesson on Parabola
Focal Chord and Focal Distance
Focal chord:Â Any chord passes through the focus of the parabola is a fixed chord of the parabola.
Focal Distance:Â The focal distance of any point p(x, y) on the parabola y^{2} = 4ax is the distance between point â€˜pâ€™ and focus.
PM = a + x
PS = Focal distance = x + a
General Equations of Parabola
Equation of parabola by definition.
SP = PM
\({{(x\alpha )}^{2}}+{{(y\beta )}^{2}}=\frac{{{(\ell x+my+n)}^{2}}}{{{\ell }^{2}}+{{m}^{2}}}\)
The general equation of 2^{nd} degree i.e. ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 if
\(\Delta \ne 0\) \({{h}^{2}}=ab\)
Position of a point with respect to parabola
For parabola
\(S\equiv {{y}^{2}}4ax=0\,\,\,\,\,\,\,\,\,,\,p(x{}_{1},{{y}_{1}})\)
\({{S}_{1}}={{y}_{1}}^{2}4a{{x}_{1}}\)
\({{S}_{1}}<0\,\,\,\,\,(inside\,curve)\)
\({{S}_{1}}=0\,\,\,\,\,(on\,curve)\)
\({{S}_{1}}>0\,\,\,\,(outside\,curve)\)
Intersection of a straight line with the parabola y^{2} = 4ax
Straight line y = mx + c
m slope of straight line
(mx + c)^{2} â€“ 4ax = 0
m^{2}x^{2} + 2x(mc â€“ 2a) + c^{2} = 0
Ax^{2} Bx + c = 0
B^{2} â€“ 4AC = discriminant D
D = 0
\(c={}^{a}/{}_{m}\)
D > 0
mc â€“ a > 0:Â Straight line intersect the curve
D < 0 (mc – a) < 0:Â Straight line not touching the curve
Tangent to a Parabola
Tangent at point (x_{1}, y_{1})
y^{2} = 4ax (parabola)
equation of Tangent
\(y{{y}_{1}}{{y}_{1}}^{2}=2a(x{{x}_{1}})\)
\(y{{y}_{1}}4a{{x}_{1}}=2a(x{{x}_{1}})\)
\(y{{y}_{1}}=2a(x+{{x}_{1}})\)
â‡’ Point \(({{x}_{1}}\,{{y}_{1}})\)
Tangent in slope (m) form:
y^{2} = 4ax
Let equation of Tangent y = mx + c
From the previous illustration
y = mx + c touches curve at a point
so , \(c\text{ }=~{}^{9}/{}_{m}\)
equation of Tangent : y = mx + \(~{}^{a}/{}_{m}\)
so, point of Tangency is \(\left( {}^{a}/{}_{{{m}^{2}}},\frac{2a}{m} \right)\)
Tangent in parameter form (at^{2}, 2at)
ty = x + at^{2 } where â€˜tâ€™ is
parameter
Pair of Tangents from (x_{1}, y_{1}) external points
Let y^{2} = 4ax, (parabola)
P(x_{1}, y_{1}) external point then equation of Tangents is given by
SS_{1} = T^{2}
\(S\equiv {{y}^{2}}4\,ax,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{1}}\equiv {{y}_{1}}^{2}4a{{x}_{1}}\)
\(T\equiv y{{y}_{1}}2a(x{{x}_{1}})\)
Chord of contact:
Equation of chord of contact of Tangents from a point p(x_{1}, y_{1}) to the parabola y^{2} = 4ax is given by T = 0
i.e., yy_{1} â€“ 2a(x + x_{1}) = 0
Equation of QS T = 0
Normal to the parabola:
Normal to the point p(x_{1}, y_{1}) since normal is perpendicular to Tangent so slope of normal be will
\({}^{1}/{}_{Slope\,of\,Tangent}\)
slope of normal at â€˜pâ€™ (x_{1} y_{1}) is \(\frac{{{y}_{1}}}{2a}\)
equation of normal\(y{{y}_{1}}=\frac{{{y}_{1}}}{2a}(x{{x}_{1}})\)
Normal in term of â€˜mâ€™:
\(\left( slope\,of\,normal \right)\Rightarrow m=\frac{dx}{dy\,\,}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}=4ax\,\,\)
\({{y}_{1}}=2am\)
\(\,\,\,\,\,\,\,m=\frac{{{y}_{1}}}{2a}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}_{1}}=a{{m}^{2}}\,\,\,\,\)
\(y=mx2ama{{m}^{3}}\)
\(m=\frac{dx}{dy}\)
Equation of normal at point (am^{2}, – 2am)
Normal at point (at^{2}, 2at)
T parameter
y = tx + 2at + at^{3}
Important Properties of Focal Chord


 If chord joining \(P=(at_{1}^{2},2a{{t}_{1}})\) and \(Q=(at_{2}^{2},2a{{t}_{2}})\)is focal chord of parabola \({{y}^{2}}=4ax\) then \({{t}_{1}}{{t}_{2}}=1\).
 If one extremity of a focal chord is \((at_{1}^{2},2a{{t}_{1}})\) then the other extremity \((at_{1}^{2},2a{{t}_{2}})\) becomes \(\left( \frac{a}{t_{1}^{2}},\frac{2a}{{{t}_{1}}} \right)\).
 If point \(P(a{{t}^{2}},2at)\) lies on parabola \({{y}^{2}}=4ax\), then the length of focal chord PQ is \(a{{(t+1/t)}^{2}}\).
 The length of the focal chord which makes an angle Î¸ with positive xaxis is \(4a\cos e{{c}^{2}}\theta\).
 Semi latus rectum is harmonic mean of SP and SQ, where P and Q are extremities of latus rectum. i.e., \(2a=\frac{2SP\times SQ}{SP+SQ}\,or\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a}\)
 Circle described on focal length as diameter touches tangent at vertex.
 Circle described on focal chord as diameter touches directrix.

Important Properties of focal chord, Tangent and normal of Parabola


 The tangent at any point P on a parabola bisects the angle between the focal chord through P and the perpendicular from P on the directix.



 The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus.

(iii) Tangents at the extremities of any focal chord intersect at right angles on the directrix.
(iv) Any Tangent to a parabola and perpendicular on it from the focus meet on the Tangent at its vertex.
Intersect at yaxis, at u = 0
Four Common Forms of a Parabola:
Form:  y^{2} = 4ax  y^{2} = – 4ax  x^{2} = 4ay  x^{2} = – 4ay 

Vertex:  (0, 0)  (0,0)  (0, 0)  (0, 0) 
Focus:  (a, 0)  (a, 0)  (0, a)  (0, a) 
Equation of the directrix:  x = – a  x = a  y = – a  y = a 
Equation of the axis:  y = 0  y = 0  x = 0  x = 0 
Tangent at the vertex:  x = 0  x = 0  y = 0  y = 0 
Practice Problems on Parabola
Illustration 1: Find the vertex, axis, directrix, tangent at the vertex and the length of the latus rectum of the parabola \(2{{y}^{2}}+3y4x3=0\).
Solution: The given equation can be rewritten as \({{\left( y+\frac{3}{4} \right)}^{2}}=2\left( x+\frac{33}{32} \right)\)
which is of the form \({{Y}^{2}}=4aX\)where \(Y=y+\frac{3}{4},\,X=x+\frac{33}{32},\,4a=2\).
Hence the vertex is \(X=0,Y=0\) i.e. \(\left( \frac{33}{32},\frac{3}{4} \right)\).
The axis is \(y+\frac{3}{4}=0\Rightarrow y=\frac{3}{4}\).
The directix is \(X=a=0\)
\(\Rightarrow x+\frac{33}{32}+\frac{1}{2}=0\Rightarrow x=\frac{49}{32}\)
The tangent at the vertex is \(X=0\,or\,x+\frac{33}{32}=0\Rightarrow x=\frac{33}{32}\).
Length of the latus rectum = 4a = 2.
Illustration 2: Find the equation of the parabola whose focus is (3, 4) and directix x â€“ y + 5 = 0.
Solution: Let P(x, y) be any point on the parabola. Then
\(\sqrt{{{(x3)}^{3}}+{{(y+4)}^{2}}}=\frac{\left xy+5 \right}{\sqrt{1+1}}\)
\(\Rightarrow {{(x3)}^{2}}+{{(y+4)}^{2}}=\frac{{{(xy+5)}^{2}}}{2}\)
\(\Rightarrow {{x}^{2}}+{{y}^{2}}+2xy22x+26y+25=0\)
\(\Rightarrow {{(x+y)}^{2}}=22x26y25\)
Illustration 3: Find the equation of the parabola having focus (6, 6) and vertex (2, 2).
Solution: Let S(6, 6) be the focus and A(2, 2) the vertex of the parabola. On SA take a point K (x_{1 }, y_{1}) such that SA = AK. Draw KM perpendicular on SK. Then KM is the directix of the parabola.
Since A bisects SK, \(\left( \frac{6+{{x}_{1}}}{2},\frac{6+{{y}_{1}}}{2} \right)=(2,2)\)
\(\Rightarrow 6+{{x}_{1}}=4\,and\,6+{{y}_{1}}=4\,or\,({{x}_{1}},{{y}_{1}})=(2,10).\)
Hence the equation of the directrix KM is y â€“ 10 = m(x+2) â€¦â€¦(1)
Also gradient of \(SK=\frac{10(6)}{2(6)}=\frac{16}{8}=2;\,m=\frac{1}{2}\)
So that equation (1) becomes
\(y10=\frac{1}{2}(x2)\) or \(x+2y22=0\) is the directrix.
Next, let PM be a perpendicular on the directrix KM from any point P(x, y) on the parabola.
From SP = PM, the equation of the parabola is
\(\sqrt{\left\{ {{(x+6)}^{2}}+{{(y+6)}^{2}} \right\}}=\frac{x+2y22}{\sqrt{({{1}^{2}}+{{2}^{2}})}}\)
Illustration 4: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for \({{y}^{2}}=12x\).
Solution: The given equation is \({{y}^{2}}=12x\).
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with \({{y}^{2}}=4ax\), we get \(4a=12a\) or \(a=3\).
Coordinates of the focus are given by (a, 0) i.e., (3, 0).
Since the given equation involves y^{2}, the axis of the parabola is the yaxis.
Equation of directix is \(x=a\), i.e., \(x=3\).
Length of latus rectum = 4a = 4 x 3 = 12.
Illustration 5: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for \({{x}^{2}}=16y\).
Solution: The given equation is \({{x}^{2}}=16y\).
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with \({{x}^{2}}=4ay\), we get \(4a=16\) or \(a=4\).
Coordinates of the focus = (0, a) = (0, 4).
Since the given equation involves \({{x}^{2}}\), the axis of the parabola is the yaxis.
Equation of directrix, y = a i.e. = 4.
Length of latus rectum = 4a = 16.
Illustration 6: If the parabola y^{2 } = 4x and x^{2} = 32y intersect at (16, 8) at an angle Î¸, then find the value of Î¸.
Solution: The slope of the tangent to y^{2} = 4x at (16, 8) is given by
\({m}_{1}={\left( \frac{dy}{dx} \right)}_{(16,8)}={{\left( \frac{4}{2y} \right)}_{(16,8)}}=\frac{2}{8}=\frac{1}{4}\)
The slope of the tangent to x^{2} = 32y at (16, 8) is given by
\({m}_{2}={\left( \frac{dy}{dx} \right)}_{(16,8)} ={{\left( \frac{2x}{32} \right)}_{(16,8)}}=1\)
âˆ´ \(Tan \;\theta =\frac{1(1/4)}{1+(1/4)}=\frac{3}{5}\)
\(\Rightarrow \,\,\,\,\,\theta ={{\tan }^{1}}\left( \frac{3}{5} \right)\)
Illustration 7: Find the equation of common tangent of y^{2} = 4ax and x^{2} â€“ 4ay.
Solution: Equation of tangent to y^{2} = 4ax having slope m is \(y=mx+\frac{a}{m}\).
It will touch x^{2} â€“ 4ay, if \({{x}^{2}}=4a\left( mx+\frac{a}{m} \right)\) has a equal roots.
Thus, \(16{{a}^{2}}{{m}^{2}}=\text{ }16\frac{{{a}^{2}}}{m}\,\,\,\Rightarrow \,m=1\)
Thus, common tangent is y + x + a = 0.
Illustration 8: Find the equation of normal to the parabola y^{2} = 4x passing through the point (15, 12).
Solution: Equation of the normal having slope m is
\(y=mx2m{{m}^{3}}\)
If it passes through the point (15, 12) then
\(12=15m2m{{m}^{3}}\)
\(\Rightarrow \,\,\,\,\,{{m}^{3}}13m+12=0\)
\(\Rightarrow \,\,\,\,\,\left( m1 \right)\left( m3 \right)\left( m+4 \right)=0\)
\(\Rightarrow \,\,\,\,\,m=1,\,3,\,4\)
Hence, equations of normal are:
\(y=x3,\,y=3x33\,and\,y+4x=72\)
Illustration 9: Find the point on y^{2} = 8x where line x + y = 6 is a normal.
Solution: Slope m of the normal x + y = 6 is 1 and a = 2
Normal to parabola at point (am^{2}, 2am) is
\(y=mx2ama{{m}^{3}}\)
\(\Rightarrow \,\,\,\,\,y=x+4+2\,at\,(2,4)\)
\(\Rightarrow \,\,\,\,\,x+y=6\,is\,normal\,at\,(2,4)\)
Illustration 10: Tangents are drawn to y^{2} = 4ax at point where the line lx + my + n = 0 meets this parabola. Find the intersection of these tangents.
Solution: Let the tangents intersects at P (h, k). Then lx + my + n = 0 will be the chord of contact. That means lx + my + n = 0 and yk â€“ 2ax â€“ 2ah = 0 which is chord of contact, will represent the same line.
Comparing the ratios of coefficients, we get
\(\frac{k}{m}=\frac{2a}{l}=\frac{2ah}{n}\)
\(\Rightarrow \,\,\,\,\,h=\frac{n}{l},\,k=\frac{2am}{l}\)
Illustration 11: If the chord of contact of tangents from a point P to the parabola If the chord of contact of tangents from a point P to the parabola y^{2} = 4ax touches the parabola x^{2}=4by, then find the locus of P.
Solution: Chord of contact of parabola y^{2} = 4ax w.r.t. point P(x_{1} , y_{1})
yy_{1} = 2a(x + x_{1}) â€¦â€¦(1)
This line touches the parabola x^{2} = 4by.
Solving line (1) with parabola, we have
\({{x}^{2}}=4b\left[ \left( 2a/{{y}_{1}} \right)\left( x+{{x}_{1}} \right) \right]\)
or \({{y}_{1}}{{x}^{2}}8abx8ab{{x}_{1}}=0\)
According to the question, this equation must have equal roots.
\(\Rightarrow \,\,\,\,\,D=0\,\)
\(\Rightarrow \,\,\,\,64{{a}^{2}}{{b}^{2}}+32ab{{x}_{1}}{{y}_{1}}=0\)
\(\Rightarrow \,\,\,\,\,{{x}_{1}}{{y}_{1}}=2ab\) or \(xy=2ab\), which is the required locus.
good and best to get suceed in a chapter