Solved IGNOU Assignments and Question papers: CURVE TRACING EXAMPLE 6

Given equation of the curve is

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

Symmetry

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

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

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

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

${y^2} = {\left( { - x} \right)^2}\left( {\frac{{a - \left( { - x} \right)}}{{a + \left( { - x} \right)}}} \right)$
$\Rightarrow {y^2} = {x^2}\left( {\frac{{a + x}}{{a - x}}} \right)$
which is same as the equation of the given curve.
So the curve is not symmetric about the y-axis and neither about the opposite quadrants.

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

${y^2} = {x^2}\left( {\frac{{a + x}}{{a - x}}} \right)$
$\Rightarrow {x^2} = {y^2}\left( {\frac{{a + y}}{{a - y}}} \right)$
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 and -x along with -y and x in the equation of the given curve.

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

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) instead of (0,0) in the equation of the given curve.

${0^2} = {0^2}\left( {\frac{{a + 0}}{{a - 0}}} \right)$
$\Rightarrow 0 = 0$ .______________________(1)

The given equation of the curve can also be written as

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

Equating the lowest degree terms to zero

${x^2} - {y^2} = 0$
$\left( {x - y} \right)\left( {x + y} \right) = 0$ _______________________(2)

From (1) and (2) the curve passes through the origin and (x-y=0) and (x+y=0) are the tangents thereat.

Point of Intersection with the x- axis and the y-axis

Put x=0 in the equation

${x^2}a - {y^2}a - {x^3} - {y^2}x = 0$ __________________________(3)

we get

y=0.

The curve meets y-axis at the origin only.

Putting y=0 in equation (3)

we get

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

$x = 0,a$

The curve meets x-axis at (a,0) and at origin.

Asymptotes

Equate the highest degree term of y to zero in

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

which leads us to

$\left( {x + a} \right) = 0$

is the asymptote parallel to y-axis.

Equate the highest degree term of x to zero in

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

which leads us to

-1=0

which is not true.

So there are no asymptotes parallel to the x-axis.

Domain

The region in which the curve lies is defined as follows,

$y = x\sqrt {\frac{{a - x}}{{a + x}}}$

The curve is defined in the region

$\left( { - a,a} \right]$

Intervals of increase and decrease

Given equation of the curve is

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

Differentiating,

$2y\frac{{dy}}{{dx}} = \frac{{2{a^2}x - 2a{x^2} - 2{x^3}}}{{{{\left( {a + x} \right)}^2}}}$
$2y\frac{{dy}}{{dx}} = \frac{{ - 2x\left( {{x^2} + xa - {a^2}} \right)}}{{{{\left( {a + x} \right)}^2}}}$

In the domain the curve increases in the interval,

$\left( { - a,0.618a} \right)$

and decreases in the interval

$\left( {0.618a,a} \right)$ .

Solved IGNOU Assignments and Question papers

Sunday, January 26, 2020

CURVE TRACING EXAMPLE 6

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