Friday, January 24, 2020

CURVE TRACING EXAMPLE 2

Given equation of the curve is,

{y^2} = \,\frac{{{x^2}\left( {x + a} \right)}}{{x - a}} ________________________(1)

Symmetry

1.Replace y by -y in the equation of the given curve, which gives

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

which is same as the equation of the given curve.

So the curve is symmetrical about x-axis.

2.Replace x by -x in the equation of the given curve, which gives

{y^2} = \,\frac{{{{\left( { - x} \right)}^2}\left( { - x + a} \right)}}{{\left( { - x - a} \right)}}

\Rightarrow {y^2} = \,\frac{{{x^2}\left( { - x + a} \right)}}{{ - \left( {x + a} \right)}}

which is not same as the equation of the given curve.

So the curve is not symmetrical about the y-axis and neither in opposite quadrants.

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

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

which is not same as the equation of the given curve.

So, the curve is not symmetrical about y=x.

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

{\left( { - x} \right)^2} = \,\frac{{{{\left( { - y} \right)}^2}\left( { - y + a} \right)}}{{\left( { - y - a} \right)}}

\Rightarrow {\left( { - x} \right)^2} = \,\frac{{{{\left( { - y} \right)}^2}\left( { - y + a} \right)}}{{ - \left( {y + a} \right)}}

Check for origin

 Replace (x,y) by (0,0) in (1)

\begin{array}{l} \Rightarrow {\left( { - 0} \right)^2} = \,\frac{{{{\left( { - 0} \right)}^2}\left( { - 0 + a} \right)}}{{ - \left( {0 + a} \right)}}\\ \Rightarrow 0 = 0 \end{array}

So the curve passes through the origin.

\begin{array}{l} {y^2}x - a{y^2} = a{x^2} + {x^3}\\ \Rightarrow a{y^2} + a{x^2} + {x^3} - {y^2}x = 0\\ \Rightarrow a\left( {{x^2} + {y^2}} \right) + {x^3} - {y^2}x = 0 \end{array}

Equating the lowest degree terms to zero,

{x^2} + {y^2} = 0
\Rightarrow \left( {x + \iota y} \right)\left( {x - \iota y} \right) = 0

Thus there are imaginary tangents at the origin hence origin is an isolated point.

Point of intersection with x-axis and y-axis

 Put x=0, in a{y^2} + a{x^2} + {x^3} - {y^2}x = 0

\Rightarrow a{y^2} = 0.
\Rightarrow y = 0.

The given curve cuts y-axis at the origin only.

Put y=0, in a{y^2} + a{x^2} + {x^3} - {y^2}x = 0
\Rightarrow x = 0, - a

The given curve cuts the x- axis at the origin and (-a,0).

Asymptotes

 Equate the coefficient of highest degree term of x to zero to find asymptotes parallel to x-axis
in the following equation

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

\Rightarrow \left( {a - x} \right) = 0 .
\Rightarrow x = a .

 Equate the coefficient of highest degree term of y to zero to find asymptotes parallel to y-axis in the following equation

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

1=0.

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

Put y=mx+c in a{y^2} + a{x^2} + {x^3} - {y^2}x = 0 to find oblique asymptotes

First we get the value of m= \pm 1.

When m=1, c=a

When m=-1, c=-a

Therefore (y=x+a) and (y=-x-a) are the two oblique asymptotes.

Domain

 The curve lies in the region

\left[ { - \infty , - a} \right)\, \cup \,\left( {a,\infty } \right]

Intervals of Increasing and Decreasing
{y^2} = \,\frac{{{x^2}\left( {x + a} \right)}}{{x - a}}


2y\frac{{dy}}{{dx}} = \frac{{2x\left( {{x^2} - ax - {a^2}} \right)}}{{{{\left( {x - a} \right)}^2}}}

From above the interval of decrease is

\left( { - \infty , - a} \right)\, \cup \,\left( {a,1.618a} \right)

Also the interval of increase is

\left( {1.618a,\infty } \right)























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