Saturday, January 25, 2020

CURVE TRACING EXAMPLE 4


Given curve is

{y^2}{x^2} = {x^2} - {a^2}______________(1)

Symmetry

1.Replace -y by y in (1).

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

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

which is same as the equation of the given curve.
So, the curve is symmetric about the y-axis but not in opposite quadrants.

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

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

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

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

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

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

Check for origin

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

{y^2}{x^2} = {x^2} - {a^2}
\Rightarrow {0^2}{0^2} = {0^2} - {a^2}
0 = - {a^2}

which is not true.

The curve does not pass through the origin.

Points of intersection with x-axis and y-axis

Put x=0 in

{y^2}{x^2} = {x^2} - {a^2}
\Rightarrow {y^2}{0^2} = {0^2} - {a^2}
0 = - {a^2}

which is not true.

The curve does not cut the y-axis.

Put y=0 in

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

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

Asymptotes

Equation of the curve is

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

Equating the highest degree term of x to zero in

{x^2}\left( {{y^2} - 1} \right) + {a^2} = 0

gives us

\left( {{y^2} - 1} \right)=0

y-1=0 and y+1=0 are the two asymptotes.

Domain

The domain of can be calculated as,

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


{x^2} - {a^2} \ge 0

The domain of y is

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

So the curve is defined in the region given above.

Intervals of increase and decrease

 Given equation of the curve is

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

Differentiating,we get

2y\frac{{dy}}{{dx}} = \frac{{2{a^2}}}{{{x^3}}}

The curve decreases in the interval \left( { - \infty , - a} \right] i.e. in its domain and increases in the interval \left[ {a,\infty } \right).



















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