# Parabola - 1

Introduction

Parabola is a curve of infinite extent and is not often obsened in real life as circles or shaight line are . However, the parabplaic curres are very important and have some vert good propertvies like the light raus reflected from parabolic surface passes through one foxed point. This property of parabola is luidely used in making lenses and in optics.

The brridges also take parabolic shape have been centuries lod. So, now let us start studying this interesting curre in more detail.

Parabola is locus of point which moves such that its distance from fixed point (focus) is always equat to its distance from fixed line (directrix) i.e. eccentricity, e = 1 .

The equation ax

(1) Second degrel thrms make a perfict square. or n

Let the fixed point be . and fixed line ax + by + 1 = 0 Let variable point be (h,k).

Now,according to the definition of parabola.

the distance of (h,K) from () is same as distance from line ax + by + 1 = 0.

So locus of

On simplification, we get

(a

The expression reduces to

b

So, second deqree terms make perfect squarc.

ax

Now if h

then

So, the two thing are same .

(2)

Ide ntify the locus of point which move such that its distance from given point and line is equal ?

(i) (-3,7) is the point and x + 2y - 8 = 0 is given line .

or,

=>. 5 (x

= x

= 4x

Considening 2

Now consider

So, the locus is a parabola

(ii) (-6,7) is point and x + 2y - 8 = 0 is given line .

So, the locus is stratight line itself .

(i) Vertex 0 (0,0)

(ii) Focus s(9,0)

(iii) Foot of directrix (-9,0)

(iv) Directrix x + a = 0

(v) Equation of catus rectum x = a

and length of Latus retum = 4a .

(vi) Axis y= 0.

(vii) Extremiofies of latus retum (9

The brridges also take parabolic shape have been centuries lod. So, now let us start studying this interesting curre in more detail.

Parabola is locus of point which moves such that its distance from fixed point (focus) is always equat to its distance from fixed line (directrix) i.e. eccentricity, e = 1 .

__Note__: The distance from fixed point to the distance from fixed line is calle__General second degree equation__: eccentncity.The equation ax

^{2}by + 2hxy + 2gx + 2fy + c = 0(1) Second degrel thrms make a perfict square. or n

^{2}= ab.__Why ?__Let the fixed point be . and fixed line ax + by + 1 = 0 Let variable point be (h,k).

Now,according to the definition of parabola.

the distance of (h,K) from () is same as distance from line ax + by + 1 = 0.

So locus of

On simplification, we get

(a

^{2}+ b^{2})=(ax + by + 1)^{2}The expression reduces to

b

^{2}x^{2}+ a^{2}y^{2}- 2abxy + .....linear terms .... + conjtant=(bx - ay)^{2}+ ......linear terms + constant = 0.So, second deqree terms make perfect squarc.

__Dumb Question__: Why is h^{2}= ab same as second term making a perfect squarc ?__Ans__The second degree terms in equation areax

^{2}+ by^{2}+ 2hxy .Now if h

^{2}= abthen

So, the two thing are same .

(2)

__Illustration 1__Ide ntify the locus of point which move such that its distance from given point and line is equal ?

(i) (-3,7) is the point and x + 2y - 8 = 0 is given line .

__Ans__Now according to the question, let the point whose locus is to be determined be (x,y).or,

=>. 5 (x

^{2}+ 9 + bx + y^{2}+ 49 -14y)= x

^{2}4y^{2}+ 64 -16 x - 32y +4xy= 4x

^{2}+ y^{2}- 4xy + 46x - 38y + 226 = 0Considening 2

^{0}term, 4x^{2}y^{2}- 4xy = (2x - y)^{2}which is a perfect square.Now consider

So, the locus is a parabola

(ii) (-6,7) is point and x + 2y - 8 = 0 is given line .

__Ans__. Note taht (- 6,7) satisfies the given line .So, the locus is stratight line itself .

__Standared Equation of parabola__.__Important properties__(i) Vertex 0 (0,0)

(ii) Focus s(9,0)

(iii) Foot of directrix (-9,0)

(iv) Directrix x + a = 0

(v) Equation of catus rectum x = a

and length of Latus retum = 4a .

(vi) Axis y= 0.

(vii) Extremiofies of latus retum (9

_{1}2a) & (9_{1}-2a)__Note__: Two parabolas are said to be equal if their length of latus rectii are equal .