Solved IGNOU Assignments and Question papers: CURVE TRACING EXAMPLE 7

Given equation of the curve is

${x^3} + {y^3} = 3a{x^2}$

Symmetry

1.Replacing -y by y in the equation of the given curve.

$\Rightarrow {x^3} - {y^3} = 3a{x^2}$

which is not same as the equation of the given curve.
So the curve is not symmetric about the x-axis.

2.Replacing -x by x in the equation of the given curve/

$\begin{array}{l} {\left( { - x} \right)^3} + {y^3} = 3a{\left( { - x} \right)^2}\\ - {x^3} + {y^3} = 3a{x^2} \end{array}$

which is not same as the equation of the given curve.
So the curve is not symmetric about the y-axis neither in opposite quadrants.

3.Interchange y by x and x by y in the equation of the given curve.

${y^3} + {x^3} = 3a{y^2}$

which is not same as the equation of the given curve.
So, the curve is not symmetric about the line y=x.

4. Interchange y by -x and -y and x in the equation of the given curve.

$\begin{array}{l} {\left( { - y} \right)^3} + {\left( { - x} \right)^3} = 3a{\left( { - y} \right)^2}\\ \Rightarrow - \left( {{x^3} + {y^3}} \right) = 3a{y^2} \end{array}$

which is not same as the equation of the given curve.
So, the curve is not symmetric about the line y=-x.

Check for Origin

Replace (x,y) by (0,0) in the equation

${x^3} + {y^3} = 3a{x^2}$

${0^3} + {0^3} = 3a{\left( 0 \right)^2}$

$0 = 0$ .

The curve passes through the origin.

${x^3} + {y^3} - 3a{x^2} = 0$

Equating the lowest degree term of x to zero we get,

x=0 is the tangent at the origin and origin is the node there.

Points of intersection with x-axis and the y-axis

Put y=0 in

${x^3} + {y^3} - 3a{x^2} = 0$

we get the points of intersection with the x-axis as

(3a,0) and (0,0).

The curve intersects y-axis at the origin only as if we put x=0 in,

${x^3} + {y^3} - 3a{x^2} = 0$

we get,

y=0.

Asymptotes

Given equation of the curve is

${x^3} + {y^3} - 3a{x^2} = 0$ ._____________________(3)

Equating the highest degree term of y to zero.

1=0.

which is not true.
Thus there are no asymptotes parallel to y-axis.

Equating the highest degree term of x to zero.

1=0.

which is not true.
Thus there are no asymptotes parallel to x-axis.

Put y=mx+c in (3) to find the oblique asymptote.

${x^3} + {\left( {mx + c} \right)^3} - 3a{x^2} = 0$

${x^3} + {m^3}{x^3} + 3{m^2}{x^2}c + 3mc{x^2} - 3a{x^2} = 0$ ___________(4)

Equating the coefficient of highest power of x to zero,

${m^3} + 1 = 0$

m=-1.

Equating the coefficient of next highest power of x to zero,

3c-3a=0

c=a.

So the oblique asymptote is,

y=-x+a.

Domain

The given curve is defined $\forall \,\mathbb{R}$ .

Intervals of increase and decrease

Given equation of the curve is

${y^3} = 3a{x^2} - {x^3}$

$2{y^2}\frac{{dy}}{{dx}} = 6ax - 3a{x^2}$

The critical points are x=0 and x=2

y is increasing in the interval (0,2) and decreasing in the interval $\left( { - \infty ,0} \right)\, \cup \,\left( {2,\infty } \right)$ .

Solved IGNOU Assignments and Question papers

Sunday, January 26, 2020

CURVE TRACING EXAMPLE 7

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