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Trace the curve : $y^{2}(2+x)=x^{2}(6-x)$

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1 Answer

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Toolbox:
  • The strategies for curve tracing
  • (1) Domain-The values of x for which the function is defined
  • (2)Extent -Horizontal (vertical ) extent is determined by the intervals for $x(y)$ for which the curve exists
  • (3) $x=0$ gives the y intercept and y=0 gives the x -intercept
  • (4) If $(0,0)$ satisfies the equation, the curve passes through the orgin
  • (5) Find out whether the curve is symmetrical about any line. The curve is symmetric about
  • (i) x-axis if the equation is unaltered when y is replaced by -y
  • (ii) y-axis if the equation is unaltered when x is replaced by -x
  • (iii)the orgin if the equation is unattened when x is replaced by -x abd y by -y
  • (iv) the line $y=x$ if the equation is uttered when x and y are interchanged
  • (v) the line $ y=-x$ if the eauation is unchanged if x and y are replaced by -y and -x
  • (6) Asymptotes ($\parallel$ to coordinate axes)
  • If $y \to c$ (finite) when $x \to \pm \infty$
  • (or $ x \to k $ (finite ), when $y \to \pm \infty$
  • (7)Monotsnicity : Determine the intervals for which the curve is increasing or decreasing, using the first derivative test
  • (8) Special points: Determine the intervals of concavity/convexity and the points of inflection, using the second derivative test.
Step 1:
$y^2(2+x)=x^2(6-x)$
or $y^2=\large\frac{x^2(6-x)}{(2+x)}$
Step 2:
Existence : The curve is not defined for $x > 6$ and $x \leq -2$. so it lies in the interval $ -2 < x \leq 1$
Step 3:
(ii) Symmetry: The curve is symmetric about the x-axis only
Step 4:
(iii) Asymptotes: $x=-2$ is a vertical asymplote to the curve , parallel to the y-axis
Step 5:
The curve passes through $(0,0)$ twice $(6,0)$ once, and hence a loop is forward between $(0,0)\;and \; (6,0)$
answered Aug 19, 2013 by meena.p
 
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