The research paper published by IJSER journal is about New Method to Find The Axis of a Parabola 1

ISSN 2229-5518

New Method to Find the Axis of a Parabola

Nulavai.Jaya Krishna

AbstractThe content describes a new way to find the axis of a parabola and a new derivation for the equation of the parabola.

Index Termsparabola,axis,partial differentiation,

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Theorem 1

Consider a parabola ax^2+2hxy+by^2+2gx+2fy+c=0,

Let us differentiate the equation 1 W.R.T x,
We get ax+hy+hxD+byD+g+fD=0; {d refers to (dy/dx)

then

(i)the slope of the axis is given by –(a/b)

(ii)the equation of the axis of the parabola is given by

Then D= -(ax+hy+g)/(hx+by+f ) (4)
This is the slope of the tangent at any point on the para-

ax+hy++((hf+ag)/(a+b))=0 which can be simplified as

ax+by+((fb+ga)/(a+b))=0.

Proof.

Consider a parabola of the form ax^2+2hxy+by^2+2gx+2fy+c=0 (1)
Then the equation of the parabola can be determined as follows: Differentiate equation (1) partially with respect to x,
We get ax+hy+g=0 (2)
Differentiate equation (1) partially with respect to x,
We get hx+by+f=0 (3)
Since h^2=a*b for a parabola, equations (2) and (3) are
Parallel to each other as the slopes are equal.
Consider the point of intersection of the equations (1) and (2) Put y=(-ax-g)/h in equation (1)….
Then
ax^2+2x(-ax-g)+b((-ax-g)/h)^2+2gx+2f(-ax-g)/h+c=0;
consider the x^2 coefficient ,
(a-2a+b*(a^2)/h^2)=0 since h^2=ab.
Consider the point of intersection of the equations (1) and (3)
Put y=(-hx-f)/b in equation (1)….
Then
ax^2+2hx(-hx-f )/b+b((-hx-f )/b)^2+2gx+2f(-hx-f )/b+c=0;
consider the x^2 coefficient
(a-2(h^2)/b+b*(h^2)/b^2)=0 since h^2=ab.
That means both the lines are parallel and they intersect the
parabola at only one point……..
As we cannot draw parallel tangents to a parabola,the lines
2&3 must be parallel to the axis as in the figure(1).
So the SLOPE of the axis of a parabola is (-a/h)or(-h/b).
Since h^2=ab,then (-a/h)=(-a/√(ab))= –(√a/√b).
Now the second part of the proof starts.
bola
As the vertex of the parabola the tangent is perpendicular to axis the slope of the tangent at vertex is (h/a) and let the vertex be (p,q).
From (4),
(h/a)=-(ap+hq+g)/(hp+bq+f )
Re- arranging the terms gives, (a^2+h^2)p+h(a+b)q+hf+ag=0; Put h^2=ab; (a+b)ap+h(a+b)q+hf+ag=0
It becomes ap+hq+{(hf+ag)/(a+b)}=0
This equation represents a line which is having slope -(a/h)
and passing through vertex (p,q)…
This is notihng but AXIS of the parabola……
So the general equation of the axis is ax+hy+{(hf+ag)/(a+b)}=0;
put h=√(ab).
Then ax+√(ab)y+{(√(ab).f+ag)/(a+b)}=0
Which implies that

√ax+√by+((f√b+g√a)/(a+b))=0.

As the equation of the axis of parabola.

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N.Jaya Krishna is currently pursuing bachelors degree program in compu- terscience engineering in National Institute of Technology Warangal,

INDIA, PH-91-9000744085. E-mail: jayakrishna226@gmail.com

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2 Author`s Bibilography

N.Jaya Krishna pursuing Bachelors degree in National Insti- tute Of Technology Warangal ,INDIA in computer science and engineering department.

IJSER © 2012

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International Journal of Scientific & Engineering Research Volume 3, Issue 5, May-2012 2

ISSN 2229-5518

Email id:jayakrishna226@gmail.com

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