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Conic section (parabola)

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Conic section


A conic section is the intersection of a plane and a cone. By changing the angle and location of intersection, we can produce a circle, ellipse, parabola or hyperbola; or in the special case when the plane touches the vertex: a point, line or 2 intersecting lines.


i) If a plane intersects a cone perpendicular to the axis, then the section is a circle.

ii) If a plane intersects a cone at a given Angle with the axis greater than the semi vertical angle then the section is ellipse.

iii) If an intersecting plane, not passing through the vertex is parallel to the generator of the cone, then the section is a parabola.

iv)if a plane intersects the double right cone such that the angle between the axis and the plane be less then the section is hyperbola.

OR

The locus of a point which moves in a such a way that the ratio of its distance from the fixed point (focus) to its distance from a fixed straight (directrix ) line is constant is called conic section. The constant ratio of focus to directrix is eccentricity.

A conic section in which value of eccentricity is zero is known as circle.

i.e. e=0

A conic section in which value of eccentricity is unity is known as parabola.

i.e. e =1

A conic section in which value of eccentricity is less than unity is known as ellipse.

i.e. e <1

A conic section in which value of eccentricity is greater than unity is known as hyperbola

i.e. e >1

 Parabola

The parabola is the curve formed from all the points that are equidistant from the directrix and the focus.

PF/PM=1(constant)


Focal distance: The distance of any point on parabola from the focus F is called focal distance.

Focal chord: Any chord of the parabola passing through the focus is called focal chord.PFP’ is a focal chord.

Latus Rectum: The focal chord which is perpendicular to the axis is called latus rectum.LFL’ is latus rectum.

Equation of parabola in standard form is y2=4ax (important for neb)


To find the equation of parabola in standard form we consider a parabola with vertex at origin and focus at S(a, 0), where a>0. Here axis of parabola is along the x-axis and directrix is parallel to y-axis. Since A is vertex of the parabola so we have AS=AK. Here A is origin so S and K lies on the opposite side so the coordinates of K is K(-a, 0).

Let P(x, y) be any point on the parabola. Draw the perpendicular PM on directrix to meet at M. Now, the coordinate of M is M(-a, y).

By definition of parabola,

PS=PM

Or, (x -a)2+ y2=(x+a)2 +(y-y)2

or, x2 - 2ax + a2+ y2= x+ 2ax + a2

or, y2=4ax

Different form of parabolas:



ParabolafocusaxisDirectrixvertexLatus rectum
y2=4ax
(a,0)y=0x=-a(0,0)4a
y2=-4ax
(-a,0)
y=0
x=a
(0,0)
4a
x2=4ay
(0,a)
x=0
y=-a
(0,0)
4a
x2=-4ay
(0,-a)
x=0
y=a
(0,0)
4a
(y-k)2 =4a(x-h)
(h+a,k)
y=kx=h-a(h,k)4a
(x-h)2 =4a(y-k)
(h,k+a)
x=hy=k-a(h,k)4a

Some points:

I) A point p(x1,y1) lies outside ,on or inside the parabola y2=4ax according as y12-4ax (<=>) 0.

II) parametric equations of parabola y2=4ax are x=at2,y=2at.

πΈπ‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ π‘‘π‘œ π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Žπ‘π‘œπ‘™π‘Ž 𝑦2 = 4π‘Žπ‘₯ π‘Žπ‘‘ π‘Ž π‘π‘œπ‘–π‘›π‘‘ (π‘₯1, 𝑦1)

 π‘‡β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘Žπ‘π‘œπ‘Žπ‘™π‘Ž 𝑖𝑠,

𝑦2 = 4π‘Žπ‘₯

π‘‘π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘‘π‘–π‘Žπ‘‘π‘–π‘›π‘” π‘œπ‘› π‘π‘œπ‘‘β„Ž 𝑠𝑖𝑑𝑒𝑠 π‘€π‘–π‘‘β„Ž π‘Ÿπ‘’π‘ π‘π‘’π‘π‘‘ π‘‘π‘œ π‘₯

 π‘‘𝑦/𝑑π‘₯=𝑑( 4π‘Žπ‘₯ )/𝑑π‘₯

Or,2y𝑑𝑦/dx=4a

Or, dy/dx=(2a/x)

Slope at (x1,y1) is 

Or , (dy/dx)(x1,y1)=(2a)/y1

π‘€β„Žπ‘–π‘β„Ž 𝑖𝑠 π‘‘β„Žπ‘’ π‘ π‘™π‘œπ‘π‘’ π‘œπ‘“ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ π‘‘π‘œ π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Žπ‘π‘œπ‘™π‘Ž 𝑦

2 = 4π‘Žπ‘₯ π‘Žπ‘‘ (π‘₯1, 𝑦1)

 m =2π‘Ž/𝑦1

π‘π‘œπ‘€ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ 𝑖𝑠

𝑦 βˆ’ 𝑦1=2π‘Ž/𝑦1*(x-x1)

Or, 𝑦𝑦1βˆ’ y12= 2π‘Žπ‘₯ βˆ’ 2π‘Žπ‘₯1

Or, 𝑦𝑦1βˆ’ 4π‘Žπ‘₯1= 2π‘Žπ‘₯ βˆ’ 2π‘Žπ‘₯1

Or 𝑦𝑦1 = 2π‘Ž(π‘₯ + π‘₯1)


πΆπ‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› π‘‘β„Žπ‘Žπ‘‘ π‘Ž 𝑙𝑖𝑛𝑒 𝑦 = π‘šπ‘₯ + 𝑐 𝑖𝑠 π‘Ž π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ π‘‘π‘œ π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Žπ‘π‘œπ‘™π‘Ž 𝑦2 = 4π‘Žπ‘₯. 𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ π‘‘π‘œ π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Žπ‘π‘œπ‘™π‘Ž 𝑦2 = 4π‘Žπ‘₯ 𝑖𝑛 π‘‘β„Žπ‘’ π‘ π‘™π‘œπ‘π‘’ π‘“π‘œπ‘Ÿπ‘š. π΄π‘™π‘ π‘œπ‘“π‘–π‘›π‘‘ π‘‘β„Žπ‘’ π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ π‘π‘œπ‘›π‘‘π‘Žπ‘π‘‘.

π‘ƒπ‘Ÿπ‘œπ‘œπ‘“: π‘ π‘’π‘π‘π‘œπ‘ π‘’ 𝑦 = π‘šπ‘₯ + 𝑐 𝑖𝑠 π‘Ž π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ π‘‘π‘œ π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Žπ‘π‘œπ‘™π‘Ž 𝑦2 = 4π‘Žπ‘₯.

π‘ π‘œπ‘™π‘£π‘–π‘›π‘” 𝑦 = π‘šπ‘₯ + 𝑐 π‘Žπ‘›π‘‘ 𝑦2 = 4π‘Žπ‘₯.

(π‘šπ‘₯ + 𝑐 )2 = 4π‘Žπ‘₯

 π‘œπ‘Ÿ, (π‘šx)2 + 2π‘šπ‘π‘₯ + 𝑐2 βˆ’ 4π‘Žπ‘₯ = 0

π‘œπ‘Ÿ, π‘š2π‘₯2 + 2 π‘šπ‘ βˆ’ 2π‘Ž π‘₯ + 𝑐2 = 0

π‘€β„Žπ‘–π‘β„Ž 𝑖𝑠 π‘žπ‘’π‘Žπ‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘ 𝑖𝑛 π‘₯ π‘Žπ‘›π‘‘ 𝑦 = π‘šπ‘₯ + 𝑐 𝑖𝑠 π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ π‘œπ‘“ 𝑦2= 4π‘Žπ‘₯, π‘ π‘œ π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘π‘Ÿπ‘–π‘šπ‘–π‘›π‘Žπ‘›π‘‘ 𝑖𝑠 π‘§π‘’π‘Ÿπ‘œ. 𝑖. 𝑒.π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯ π‘Žπ‘Ÿπ‘’ π‘’π‘žπ‘’π‘Žπ‘™.

4 (π‘šπ‘-a)2 βˆ’  4π‘š2c2=0

π‘œπ‘Ÿ, (π‘šc)2 βˆ’ π‘Žπ‘π‘š + π‘Ž2 βˆ’ π‘š2𝑐2 = 0

π‘œπ‘Ÿ, βˆ’4π‘Žπ‘π‘š + 4π‘Ž2= 0

π‘œπ‘Ÿ, π‘π‘š = π‘Ž

π‘œπ‘Ÿ, 𝑐 =π‘Ž/π‘š

π‘€β„Žπ‘–π‘β„Ž 𝑖𝑠 π‘‘β„Žπ‘’ π‘Ÿπ‘’π‘žπ‘’π‘Ÿπ‘’π‘‘ π‘π‘œπ‘›π‘‘π‘–π‘‘π‘–π‘œπ‘› π‘“π‘œπ‘Ÿ 𝑦 = π‘šπ‘₯ + 𝑐 π‘‘π‘œ 𝑏𝑒 π‘Ž π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘.

π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ 𝑖𝑛 π‘ π‘™π‘œπ‘π‘’ π‘“π‘œπ‘Ÿπ‘š 𝑖𝑠 𝑦 = π‘šπ‘₯ +π‘Ž/m


π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ π‘π‘œπ‘›π‘‘π‘Žπ‘π‘‘: 𝐼𝑛 π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 𝑦𝑦1 = 2π‘Ž (π‘₯ +π‘₯1) ,( π‘₯1,𝑦1) 𝑖𝑠 π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ π‘π‘œπ‘›π‘‘π‘Žπ‘π‘‘

𝑦𝑦1 = 2π‘Ž π‘₯ + π‘₯1 π‘π‘Žπ‘› 𝑏𝑒 π‘€π‘Ÿπ‘–π‘‘π‘‘π‘’π‘› π‘Žπ‘ 

2π‘Žπ‘₯ βˆ’ 𝑦𝑦1 + 2π‘Žπ‘₯1 = 0 … … . . (𝑖)

π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ 𝑖𝑛 π‘ π‘™π‘œπ‘π‘’ π‘“π‘œπ‘Ÿπ‘š 𝑖𝑠 𝑦 = π‘šπ‘₯ +a/π‘š

π‘œπ‘Ÿ, π‘šπ‘₯ βˆ’ 𝑦 +π‘Ž/m= 0 … … … 𝑖𝑖 

π‘ π‘’π‘π‘π‘œπ‘ π‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 𝑖 π‘Žπ‘›π‘‘ 𝑖𝑖 π‘Ÿπ‘’π‘π‘Ÿπ‘’π‘ π‘’π‘›π‘‘ π‘ π‘Žπ‘šπ‘’ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘

2π‘Ž/π‘š=βˆ’π‘¦1/βˆ’1=2π‘Žπ‘₯1/(π‘Ž/m)

π‘“π‘Ÿπ‘œπ‘š π‘“π‘–π‘Ÿπ‘ π‘‘ π‘‘π‘€π‘œ π‘Ÿπ‘Žπ‘‘π‘–π‘œ

2π‘Ž/π‘š=𝑦1/1

π‘œπ‘Ÿ, 𝑦1=2π‘Ž/π‘š

π‘‘π‘Žπ‘˜π‘–π‘›π‘” π‘“π‘–π‘Ÿπ‘ π‘‘ π‘Žπ‘›π‘‘ π‘™π‘Žπ‘ π‘‘ π‘Ÿπ‘Žπ‘‘π‘–π‘œ

Or,2π‘Ž/π‘š=2π‘Žπ‘šπ‘₯1/π‘Ž

π‘œπ‘Ÿ π‘₯1 =π‘Ž/π‘š2

𝐻𝑒𝑛𝑐𝑒 π‘‘β„Žπ‘’ π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ π‘π‘œπ‘›π‘‘π‘Žπ‘π‘‘ 𝑖𝑠( π‘Ž/π‘š2,2π‘Ž/π‘š)

 

Equation of the normal at the point( π‘₯1, 𝑦1) of the parabola 𝑦2 = 4π‘Žπ‘₯.

Proof: The equation of the tangent to the parabola 𝑦2 = 4π‘Žπ‘₯ at (x1, 𝑦1)
 π‘–𝑠 𝑦𝑦1 = 2π‘Ž π‘₯ + π‘₯1

slope of tangent = 2π‘Ž/𝑦1

slope of normal =βˆ’π‘¦1/2π‘Ž

the passing point of the normal is( π‘₯1, 𝑦1)

the equation of normal is

π‘¦βˆ’π‘¦1 =(βˆ’π‘¦1/2π‘Ž)(π‘₯ βˆ’ π‘₯1)

Equation of normal in slope form

the equation of normal is

π‘¦βˆ’π‘¦1 =(βˆ’π‘¦1/2π‘Ž)(π‘₯ βˆ’ π‘₯1) ………(i)

let, π‘š =βˆ’π‘¦1/2π‘Ž

or, 𝑦1 = βˆ’2π‘Žπ‘š

since (π‘₯1, 𝑦1) lies on 𝑦2= 4π‘Žπ‘₯

𝑦12 = 4π‘Žπ‘₯1

(βˆ’2π‘Žπ‘š)2 = 4π‘Žπ‘₯1

π‘₯1 = π‘Žπ‘š2

from (i)

𝑦 + 2π‘Žπ‘š = π‘š(π‘₯ βˆ’ π‘Žπ‘š2)

𝑦 = π‘šπ‘₯ βˆ’ 2π‘Žπ‘š βˆ’ π‘Žπ‘š3

which is required equation of normal in normal form.


Equation of tangent and normal to the parabola in parametric form

Proof: The equation of parabola is 𝑦2 = 4π‘Žπ‘₯

and the point of contact (π‘Žπ‘‘2, 2π‘Žπ‘‘)

Putting this point in equation yy1=2a(x-x1) we get

yο»Ώ*2at=4a*(x-at2)

Or yt=x+at(eqn of tangent)

 π‘¦ + 𝑑π‘₯ = 2π‘Žπ‘‘ + π‘Žπ‘‘(eqn of normal)