Saturday, January 25, 2020

CURVE TRACING EXAMPLE 5



Given equation of the curve is

{y^2} = \frac{{2x - 1}}{{{x^2} - 1}}______________________(1)

Symmetry

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

\begin{array}{l} {\left( { - y} \right)^2} = \frac{{2x - 1}}{{{x^2} - 1}}\\ \Rightarrow {y^2} = \frac{{2x - 1}}{{{x^2} - 1}} \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.

\begin{array}{l} {y^2} = \frac{{\left( { - 2x - 1} \right)}}{{{{\left( { - x} \right)}^2} - 1}}\\ \Rightarrow {y^2} = \frac{{ - \left( {2x + 1} \right)}}{{{x^2} - 1}} \end{array}

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

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

{y^2} = \frac{{2x - 1}}{{{x^2} - 1}}
{x^2} = \frac{{2y - 1}}{{{y^2} - 1}}

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

\begin{array}{l} {\left( { - y} \right)^2} = \frac{{2\left( { - x} \right) - 1}}{{{{\left( { - x} \right)}^2} - 1}}\\ \Rightarrow {y^2} = \frac{{ - \left( {2x + 1} \right)}}{{{x^2} - 1}} \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

Put (x,y)=(0,0) in

{y^2} = \frac{{2x - 1}}{{{x^2} - 1}}

{\left( 0 \right)^2} = \frac{{2 * \left( 0 \right) - 1}}{{{{\left( 0 \right)}^2} - 1}}
\Rightarrow 0 = 1

which is not true.

The curve does not pass through the origin.

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

 Put x=0 in

{x^2}{y^2} - {y^2} - 2x + 1 = 0

- {y^2} + 1 = 0

\left( {y + 1} \right)\left( {y - 1} \right) = 0

The curve intersects the y-axis at (0,1) and (0,-1).

Put y=0 in

{x^2}{y^2} - {y^2} - 2x + 1 = 0

- 2x + 1 = 0

The curve intersects the x-axis at \left( {\frac{1}{2},0} \right).

Asymptotes

 {x^2}{y^2} - {y^2} - 2x + 1 = 0

Equating the highest degree term of y to zero,

\Rightarrow \left( {x - 1} \right)\left( {x + 1} \right) = 0

(x-1=0) and (x+1=0) are the two asymptotes parallel to the y-axis.

Equating the highest degree term of x to zero,
y=0, is the asymptote parallel to x-axis.

For oblique asymptote put y=mx+c in
\left( {{x^2} - 1} \right){y^2} - 2x + 1 = 0
\left( {{x^2} - 1} \right){\left( {mx + c} \right)^2} - 2x + 1 = 0
\left( {{x^2} - 1} \right)\left( {{m^2}{x^2} + 2mxc + {c^2}} \right) - 2x + 1 = 0
{m^2}{x^4} + 2m{x^3}c + {c^2}{x^2} - {m^2}{x^2} - 2mcx - {c^2} - 2x - 1 = 0

So the value of m comes out to be 0 therefore there are no obique asymptotes of the given curve.

 Domain

 For domain of definition we proceed as follows,

 {y^2} = \frac{{2x - 1}}{{{x^2} - 1}}

y = \sqrt {\frac{{2x - 1}}{{{x^2} - 1}}}

y is defined for \left( { - 1,\frac{1}{2}} \right]\, \cup \,\left( {1,\infty } \right)

Intervals of increase and decrease

 Let us find intervals of increase and decrease in the region for which the curve is defined.

 {y^2} = \frac{{2x - 1}}{{{x^2} - 1}}

2y\frac{{dy}}{{dx}} = \frac{{ - 2\left( {{x^2} - x + 1} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}

In the domain \frac{{dy}}{{dx}} is decreasing for \left( { - 1,\frac{1}{2}} \right] and increasing for \left( {1,\infty } \right).




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