# Trace the curve y^2(x^2-9) = x by stating all the properties used in tracing.

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by chayannitdgp10

## Answers

2016-03-30T05:17:17+05:30

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The curve does not exist for -3<x<3 because there can be no solutions for y when x^2<9.

Because of the presence of x^2 and y^2 the curve is essentially reflected in all quadrants. If we examine QI only we have the same patterns reflected in the other quadrants. x=-3 and x=3 are asymptotic.

As x gets larger, the constant 9 becomes more insignificant and y^2 becomes x^2, in other words, the curve in QI has an asymptote y=x, which is reflected in all the other quadrants. So we know the curve is trapped between the vertical asymptote and the line y=x.

(2ydy/dx)(x^2-9)+2y^2x=4x^3 is the derivative.

When dy/dx=0, there is a turning point. y^2=2x^2.

That is, x^4/(x^2-9)=2x^2. x^4=2x^4-18x^2; x^2=2x^2-18; x^2=18, x=3sqrt(2) and y^2=36, so y=6. In QI then, there is a turning point at (3sqrt(2),6). This is reflected in each quadrant.

This is a minimum (in QI) because the curve is trapped between the above asymptotic.

There is enough information now to trace the curve(s).

The asymptotic resemble a cross intersected by vertical lines.

The asymptotic in QI and QIV has been drawn in to illustrate how close the curve comes to it.