Solved IGNOU Assignments and Question papers: CURVE TRACING EXAMPLE 1

Given equation of the curve is
${y^2} = {x^2}\left( {\frac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} \right)$ ------------------------------(1)
SYMMETRY

Replace y by -y in (1).

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

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

which is same as the equation of the given curve so the curve is symmetric about x-axis.

Replace x by -x in (1).

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

which is same as the equation of the given curve so the curve is symmetric about y-axis.

Interchange x and y in (1)

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

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

Interchange y by -x and x by -y in (1)

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

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

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

Consider

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

Replace x by -x in f(x)

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

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

Thus f(x)=f(-x) and not f(x)=-f(-x) .

CHECK FOR ORIGIN

Put (x,y)=(0,0) in the equation of the given curve.

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

$\Rightarrow 0 = 0$ .

Thus the curve passes through the origin.

The given equation of the curve can also be written as

${a^2}{y^2} + {x^2}{y^2} + {x^4} - {a^2}{x^2} = 0$ .

${a^2}({y^2} - {x^2}) + {x^2}{y^2} + {x^4} = 0$ .

Equating the lowest degree terms to zero.

y-x=0 and x+y=0 are the tangents are the tangents at the origin thereby origin is a node.

POINTS OF INTERSECTION WITH X AND Y-AXIS

Put x=0 in the equation of the given curve we get y=0.

The curve intersects the y- axis at origin.

Put y=0 in the equation of the given curve we get $x = \pm a$ .

The curve intersects the x- axis at x=a and x=-a.

ASYMPTOTES

The given equation of the curve can also be written as

${a^2}{y^2} + {x^2}{y^2} + {x^4} - {a^2}{x^2} = 0$ .-------------------(2)

${a^2}({y^2} - {x^2}) + {x^2}{y^2} + {x^4} = 0$ .

Equating the highest degree term of y to zero we get 1=0. There are no asymptotes parallel to x-axis.

Equating the highest degree term of x to zero we get ${x^2} + {a^2} = 0$ . There are no asymptotes parallel to y-axis as well as x has imaginary roots here.

Putting y=mx+c in (2) will not yield us oblique asymptotes as highest power of y is 2 and x is 4.

DOMAIN

The domain of the function is [-a,a].

INCREASING AND DECREASING

Given equation of the curve is

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

Taking its derivative,

$2y\frac{{dy}}{{dx}} = 2x\left[ {\frac{{ - {x^4} - 2{x^2}{a^2} + {a^4}}}{{{{\left( {{a^2} + {x^2}} \right)}^2}}}} \right]$

Evaluating the numerator to zero

$- {x^4} - 2{x^2}{a^2} + {a^4} = 0$

This will give us

$x = \pm 0.66a$

In the domain the given curve is increasing in the interval

$\left( { - 0.66a,0.66a} \right)$

and decreasing in the interval

$\left( { - a, - 0.66a} \right) \cup \left( {0.66a,a} \right)$

Solved IGNOU Assignments and Question papers

Wednesday, January 15, 2020

CURVE TRACING EXAMPLE 1

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