Conic Section

The names parabola and hyperbola are given by Apolonius. These curve are infact, known as conelike subdivisions or more normally conics because they can be obtained as intersections of a plane with a dual brushed right handbill cone. These curves have a really broad scope of applications in Fieldss such as planetal gesture, design of telescope and aerial, reflectors in flash visible radiation and cars headlamps etc.

Section of a cone: –

Let cubic decimeter be a fixed perpendicular line and ‘m ‘ be another line crossing it at a fixed point ‘v ‘ and inclined to it at an angle ? .

Suppose we rotate the line ‘m ‘ around line 1 in such a manner that the angle ? remains changeless. Then the surface generated is a doubly-napped right handbill excavate cone.

The point V is called the vertex ; the line 1 is the axis of the cone. The revolving line ‘m ‘ is called a generator of the cone. The vertex separates the cone into two parts called nappes.

If we take the intersection of a plane with a cone, the subdivision so obtained is called conelike subdivision. Therefore, conelike subdivisions are the curve obtained by crossing the right handbill cone by a plane.

We obtain different type of conelike subdivision depending on the place of the crossing plane with regard to the cone and by the angle made by the crossing plane with the perpendicular axis of the cone.

The intersect of the plane with the cone can take topographic point either at vertex or any other portion of the nappe either below or above the vertex.

Circle, Ellipse, Parabola, Hyperbola

When the plane cuts the nappe of the cone, we have the undermentioned state of affairs –

1. When ? = 90 grade, the subdivision is circle.
2. ? & lt ; ? & lt ; 90 grade, the subdivision is ellipse.
3. ? = ? , the subdivision is parabola.

( In each of above three state of affairs, the plane cuts wholly across on enappe of the cone )

vitamin D ) 0 ? ? & lt ; ? ; the plane cuts through both the

Nappe & A ; the curve of intersection is a hyperbola. Nappe & A ; the curve of intersection is a hyperbola.

Degenerated conelike subdivision

When the plane cuts at the vertex of the cone, we have the undermentioned different instances.

1. When ? & lt ; ? ? 90 grade, the subdivision is a point.
2. When ? = ? , the plane contains the generator of the cone & A ; the subdivision is a consecutive line. ( It is the degenerated instance of parabola )
3. When 0 ? ? & lt ; ? , the subdivision is a brace of crossing consecutive lines.

( It is degenerated signifier of a hyperbola ) .

Summary of Basic Properties

Or

	Circle	Ellipse	Parabola	Hyperbola
Standard Cartesian Equation:	x2 + y2 = r2	y2 = 4ax
Eccentricity ( vitamin E ) :	0	0 & lt ; e & lt ; 1	1	1 & lt ; e
Relation between a, B and vitamin E	B = a	b2 = a2 ( 1-e2 )		b2 = a2 ( e2-1 )
Parametric Representation		ten = at2 Y = 2at
Definition: It is the venue of all points which meet the status…	distance to the beginning is changeless	amount of distances to each focal point is changeless	distance to concentrate = distance to directrix	difference between distances to each focal point is changeless

It might clean up the logic up to see a circle to be a particular instance of an oval. Then there are two ‘main ‘ categories

• an oval, with vitamin E & lt ; 1
• a hyperbola, with vitamin E & gt ; 1

And a ‘critical ‘ category – the parabola with e = 1.

The General Equation of a Conic

The General Equation for a Conic is

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

The existent type of conic can be found from the mark of B2 – 4AC

If B2 – 4AC is…	Then the curve is a…
& lt ; 0	Ellipse, circle, point or no curve
= 0	Parabola, 2 parallel lines, 1 line or no curve.
& gt ; 0	& gt ; 0

Polar Form

For an beginning at a focal point, the general polar signifier ( apart from a circle ) is

where L is the semi latus rectum.

Ellipse

The Cartesian equation of an oval is

where a and B would give the lengths of the semi-major and semi-minor axes.

In its general signifier, with the beginning at the centre of co-ordinates

• the focal point are at
• the directrix are at
• the major axis of length 2a
• the minor axis is of length 2b
• the semi latus rectum is of length

From the general polar signifier, the equation for an oval is

For any point P on the margin, the amount

PF1 + PF2

will be changeless, no affair which point is chosen as P.

Hence, an oval can besides be defined as the venue of a point which moves in a plane so that the amount of its distances from two fixed points is changeless. Harmonizing to Kepler ‘s First jurisprudence, the orbit of a planet is an oval. The Earth is shaped like an ellipsoid.

Any signal from one of the focal point will go through thru the other focal point.

Hyperbola

The Cartesian equation of an hyperbola is

In its general signifier, with the beginning at the centre of co-ordinates

• the focal point are at ( +/- ae, 0 )
• the directrix are at x = +/- a/e
• the cross axis of length 2a
• the conjugate axis is of length 2b
• the semi latus rectum is of length 2b2/a

Note the similarity in notation with eclipsiss ; although now the eccentricity is greater than one

Besides by analogy with an oval

For any point P on a hyperbola, the amount

PF1 – PF2

will be changeless, no affair which point is chosen as P.

Hence, a hyperbola can besides be defined as the venue of a point which moves in a plane so that the difference of its distances from two fixed points is changeless.

Asymptotes of Hyperbola

Rejigging the hyperbola expression to

As ten becomes larger, y tends to

These are the equations of the asymptotes.

Similar & A ; Diagonalizable Matrixs

Let A and B be square matrices of the same order.The matrix A is said to be similar to the matrix B if their exists an invertible matrix P such that

A=P-1BP or PA=BP — — — — ( 1 )

Post multiplying eqn ( 1 ) by p-1, we get

PAP-1 =B

Therefore A is similar to B if and merely if B is similar to A. The matrix P is called similarity matrix.

Diagonalizable matrices

A matrix A is diagonalizable, if it is similar to diagonal matrix, that is there exist an invertible matrix P such that P-1AP =D, where D is a diagonal matrix. Since similar matrix has the same Eigen values, the diagonal component of D are the characteristic root of a square matrixs of A. A necessary and sufficient status for the being of P is given by the undermentioned theorem.

Theorem: A square matrix of order N is diagonalizable if and merely if it has n linearly independent eigenvectors.

Quadratic Forms –

Let x = ( x1, x2, ……….xn ) be an arbitary vector in IRn. a existent quadratic signifier is an homogenous look of the signifier

Q = ?ni=1 ?nj=1 aijxixj

In which the sum power in each term is 2. Expanding, we can compose

Q = a11x2 + ( a12 + a21 ) x1x2 + ……….+ ( a1n +an2 ) x2xn +….+annx2n

= xTAx

Using the definition of matrix generation. Now, set bij = ( aij + aji ) /2.

The matrix B = ( bij ) is symmetric since bij =bji. Further, bij+bji = aij + aji.

Hence,

Q = xTBx

Where B is a symmetric matrix and bij = ( aij + aji ) /2.

For illustration, for n = 2, we have

B11 =a11, b12 = ( a12 + a21 ) /2 and b22 = a22.

Ques. — Find out what type of conelike subdivision the undermentioned quadratic signifier represents and transform it to principal axes

Q=17×12-30x1x2+17×22=128

We have Q = xT Ax, where,

This gives the characteristic equation ( 17- ? ) 2 – 152 = 0.

It has the root ?1= 2, ?2= 32.

Hence it becomes,

Q = 2y12 + 32y22

We see that Q = 128, represents the oval.

2y12 + 32y22 = 128, i.e.

y12/82 + y22/22 = 1

If we want to cognize the way of the principal axes in the x1x2- co-ordinate, we have to find normalized Eigen vector from ( A- ?I ) X=0 with

? = ?1= 2, & A ; ? = ?2 = 32 & A ; so, we get,

& A ;

Hence,

X=Xy =

X1 = y1/2 – y2/2

X2= y1/2 + y2/2

This is 450 rotary motion.