Answer to Question #136001 in Calculus for Prerna Singh

Question

trace the curve y^2(1+x^2)=x^2(1-x^2) stating all the properties used

Answer 1

$y^2 (1 + x^2) = x^2(1 - x^2)\\ \textsf{Differentiating both sides, we have;}\\ \begin{aligned} 2yy' + 2yy'x^2 + 2xy^2 &= 2x - 4x^3\\ y'(2y + 2yx^2) + 2xy^2 &= 2x - 4x^3 \end{aligned}\\ \textsf{At stationary point}\hspace{0.1cm} y' = 0.\\ \begin{aligned} y^2 &= \frac{2x - 4x^3}{2x}\\ y^2 &= 1 - 2x^2\\ x^2(1 - x^2) &= (1 - 2x^2)(1 + x^2)\\ x^2 - x^4 &= 1 + x^2 - 2x^2 - 2x^4\\ x^4 &= -2x^2 + 1\\ x^4 + 2x^2 - 1 &= 0 \end{aligned} \\ x^2 = \frac{-2 \pm \sqrt{8}}{2}\\ x^2 = -1 \pm \sqrt{2}\\ \textsf{The RHS of}\hspace{0.1cm} x^2 \textsf{cannot be negative}\\\textsf{in order to yield} \hspace{0.1cm}x\\ \therefore x = \pm\sqrt{\frac{-2 + \sqrt{8}}{2}} = \pm\sqrt{{-1 + \sqrt{2}}}\\ \therefore x = \pm\sqrt{\frac{-2 + \sqrt{8}}{2}} = \pm\sqrt{{-1 + \sqrt{2}}}\\ x = \pm \sqrt{{-1 + \sqrt{2}}} \approx \pm 0.6436\\ \textsf{At}\hspace{0.1cm} x = 0.6436, y = \pm 0.6436\sqrt{\frac{1 - 0.6436^2}{1 + 0.6436^2}}, y \approx \pm 0.4142.\\ \textsf{At}\hspace{0.1cm} x = -0.6436, y \approx \mp 0.4142 \\ \therefore \textsf{The stationary points are}\\(0.6436, 0.4142), (0.6436, -0.4142), \\(-0.6436, 0.4142),(-0.6436, -0.4142).\\ \textsf{Calculating the}\hspace{0.1cm}x \hspace{0.1cm} \textsf{and} \hspace{0.1cm}y\hspace{0.1cm}\textrm{intercepts}\\ \textsf{At}\hspace{0.1cm} x = 0, y = 0 \\ \textsf{At} \hspace{0.1cm}y = 0, x\sqrt{1 - x²} = 0 \Rightarrow x = 0, x = -1, 1. \\ \therefore \textsf{The}\hspace{0.1cm}x \hspace{0.1cm}\textrm{intercept}\hspace{0.1cm} \textsf{is}\hspace{0.1cm} (-1, 0),(0,0) \hspace{0.1cm}\& \hspace{0.1cm}(1, 0)\\ \textsf{and}\hspace{0.1cm}\textsf{the}\hspace{0.1cm} y\hspace{0.1cm}\textrm{intercept} \hspace{0.1cm} \textsf{is} \hspace{0.1cm} (0, 0). \\ \textsf{The graph is in the given below}$