TRACING OF CARTESIAN CURVES

By Kaushal Patel

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Symmetry The curve is symmetric about x-axis if the power

of y occurring in the equation are all even, i.e. f(x,-y)=f(x, y).

The curve is symmetric about y-axis if the powers of x occurring in equation are all even, i.e. f(-x, y)=f(x, y).

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Continue… The curve is symmetric about the line y= x, if on

interchanging x and y, the equation remains unchanged, i.e. f(y, x)=f(x, y).

The curve is symmetric in opposite quadrants or about origin if on replacing x by –x by y by –y, the equation remains unchanged, i.e. f(-x,-y)=f(x, y).

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Origin The curve passes through the origin if there is no

constant term in the equation. If curve passes through the origin, the tangents at the

origin are obtained by equating the lowest degree term in x and y to zero.

If there are two or more tangents at the origin, it is called a node, a cusp or an isolated point if the tangents at this point are real and distinct, real and coincident or imaginary respectively.

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Point of intersection The point of intersection of curve with x and y

axis are obtained by putting y=0 and x=0 respectively in the equation of the curve.

Tangent at the point of intersection is obtained by shifting the origin to this point and then equating the lowest degree term to zero.

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Special Points Cusps: If tangents are real and coincident then the double point

is called cusp.

Nodes: If the tangents are real and distinct then the double point is called node.

Isolated Point: If the tangents are imaginary then double point is called isolated point.

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Asymptoes Asymptotes parallel to x-axis are obtain by

equating the coefficient of highest degree term of the equation to zero.

Asymptotes parallel to y-axis are obtained by equating the coefficient of highest degree term of y in the equation to zero.

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Continue… Oblique asymptotes are obtained by the following method:Let y= mx + c is the asymptote to the curve and 2(x, y),3(x,

y) are the second and third degree terms in equation.Putting x=1 and y=m in 2(x, y) and 3(x, y) 2(x, y)=2(1, m) or 2(m) 3(x, y)=3(1, m) or 3(m) Find c=-(2(m)/’3(m))Solve 3(m)=0 m=m1,m2,……

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Region of Existence This region is obtained by expressing one variable in

terms of other, i.e., y=f(x)[or x=f(y)] and then finding the values of x (or y) at which y(or x) becomes imaginary. The curve does not exist in the region which lies between these values of x (or y).

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•Trace the cissoid y²(2a-x)=x³.1. Symmetry: The power of y in the equation of curve

is even so the curve is symmetric about x- axis.2. Origin: The equation of curve dose not contain any

constant term so the curve passes through the origin. to find tangents at the origin equating lowest degree term to zero, 2ay²= 0 => y²= 0 => y= 0 Thus x-axis be a tangent.

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Continue…3. Points of intersection: Putting y=0, we get x=0. Thus,

the curve meets the coordinate axes only at the origin.

4. Asymptotes: a) Since coefficient of highest power of is constant, there is

no parallel asymptote to x- axis.b) Equating the coefficient of highest degree term of y to

zero, we get 2a-x= 0 =>x= 2a is the asymptote parallel to y-axis.

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Continue…

5. Region:

We can write the equation of curve like

y²= x³ (2a- x) The value of y

becomes imaginary when x<0 or x>2a.

Therefore, the curve exist in the region 0<x<2a.

y

Ox

x= 2a

Cusp

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•Trace the witch of agnesi xy²= 4a²(a- x)

1. Symmetry: The powers of y is even in equation of curve so curve is symmetric about x- axis.

2. Origin: The equation of curve contain a constant term so curve dose not passes through origin.

3. Point of intersection: Putting y=0, we get x= a. Thus the curve meets the x- axis at (a, 0).

to get tangent at (a, 0), Shifting the origin to (a, 0). x=a be tangent at the origin.

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Continue…4. Asymptotes:a) Equating the coefficient of highest power of x to

zero, we get y² +4a² = 0 which gives imaginary values. Thus, there is no asymptote parallel to x-axis.

b) Equating the coefficient of highest power of y to zero, we get x = 0. Thus, y-axis be the asymptote.

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Continue…5. Region:From the equation of

curve,

y² = 4a²(a – x) x The equation becomes

imaginary when x<0 or x>a.

Therefore the curve lies in the region 0<x<a.

A(a, 0)

x = a

y

O x

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•Trace the curve (x²-a²)(y²-b²)=a²b²

1. Symmetry: The powers of x and y both are even so the curve is symmetric about both x and y-axis.

2. Origin: By putting point (0, 0) in equation, equation is satisfied. So, the curve is passes through the origin.

to find tangents at the origin equating lowest degree

term to zero, -b²x²-a²y² = 0 we get imaginary values. So, the tangents at the origin

are imaginary.

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Continue…3. Point of Intersection and Special Points:a) Tangents at the origin are imaginary. So, the origin is

isolated point.b) Curve dose not meet x and y axes.4. Asymptotes:a) Equating the coefficient of highest power of x to zero,

we get y²-b² = 0. Thus y=±b are the asymptotes parallel to x- axis.

b) Equating the coefficient of highest power of y to zero, we get x²-a² = 0. Thus, x=±a are the asymptotes parallel to y-axis.

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Continue…5. Region: From the equation of

curve, y=± bx and x=±

ay √x² -a²

√y² -b²

y is imaginary when –a<x<a and x is imaginary when –b<y<b.

Therefore the curve lies in the region -∞<x<-a, a<x<∞ and -∞<y<-b, b<y<∞.

y

xO

x= -a

x= a

y= b

y= -b

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•Trace the curve y=x²+2x x²+4

1. Symmetry: The equation of curve is not satisfy any condition of symmetricity. So, curve is not symmetric.

2. Origin: The equation of curve dose not contain any constant term. So, curve passes through the origin.

Equating lowest degree term to zero, 4y-2x=0 =>x=2y is tangent at the origin.

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Continue…3. Point of intersection and Special points: Putting y=0, we get x=0, -2. Thus, the curve meets the

x-axis at A(-2, 0) and o(0, 0). Shifting the origin to P(-2, 0) by putting x=X-2, y=Y+0 in the equation of the curve,

Y(X-2)²+4 = (X-2)²+2(X-2) =>Y(X²-4X+8)=X²+6X Equating lowest degree term to zero, 8Y-6X=0 =>4y-3x-6=0 is a tangent at (-2, 0). Curve dose not contain any special points.

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Continue…4. Asymptotes:a) Equating the coefficient of highest degree of x to zero,

we get y-1=0 Thus y=1 is asymptote parallel to x-axis.b) Equating the coefficient of highest power of y to zero,

we get x² + 4 = 0 which gives imaginary number . Thus, there is no asymptote parallel to y-axis.

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Continue…5. Region: y is define for all values of x. Thus, the

curve lies in the region -∞<x<∞.

B(2, 1)

OA(-2, 0)

y

x

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References… www.tatamcgrawhill.com

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Thank you