y 2 ( 1 + x 2 ) = x 2 ( 1 − x 2 ) Differentiating both sides, we have; 2 y y ′ + 2 y y ′ x 2 + 2 x y 2 = 2 x − 4 x 3 y ′ ( 2 y + 2 y x 2 ) + 2 x y 2 = 2 x − 4 x 3 At stationary point y ′ = 0. y 2 = 2 x − 4 x 3 2 x y 2 = 1 − 2 x 2 x 2 ( 1 − x 2 ) = ( 1 − 2 x 2 ) ( 1 + x 2 ) x 2 − x 4 = 1 + x 2 − 2 x 2 − 2 x 4 x 4 = − 2 x 2 + 1 x 4 + 2 x 2 − 1 = 0 x 2 = − 2 ± 8 2 x 2 = − 1 ± 2 The RHS of x 2 cannot be negative in order to yield x ∴ x = ± − 2 + 8 2 = ± − 1 + 2 ∴ x = ± − 2 + 8 2 = ± − 1 + 2 x = ± − 1 + 2 ≈ ± 0.6436 At x = 0.6436 , y = ± 0.6436 1 − 0.643 6 2 1 + 0.643 6 2 , y ≈ ± 0.4142. At x = − 0.6436 , y ≈ ∓ 0.4142 ∴ The stationary points are ( 0.6436 , 0.4142 ) , ( 0.6436 , − 0.4142 ) , ( − 0.6436 , 0.4142 ) , ( − 0.6436 , − 0.4142 ) . Calculating the x and y intercepts At x = 0 , y = 0 At y = 0 , x 1 − x 2 = 0 ⇒ x = 0 , x = − 1 , 1. ∴ The x intercept is ( − 1 , 0 ) , ( 0 , 0 ) & ( 1 , 0 ) and the y intercept is ( 0 , 0 ) . The graph is in the given below 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} y 2 ( 1 + x 2 ) = x 2 ( 1 − x 2 ) Differentiating both sides, we have; 2 y y ′ + 2 y y ′ x 2 + 2 x y 2 y ′ ( 2 y + 2 y x 2 ) + 2 x y 2 = 2 x − 4 x 3 = 2 x − 4 x 3 At stationary point y ′ = 0. y 2 y 2 x 2 ( 1 − x 2 ) x 2 − x 4 x 4 x 4 + 2 x 2 − 1 = 2 x 2 x − 4 x 3 = 1 − 2 x 2 = ( 1 − 2 x 2 ) ( 1 + x 2 ) = 1 + x 2 − 2 x 2 − 2 x 4 = − 2 x 2 + 1 = 0 x 2 = 2 − 2 ± 8 x 2 = − 1 ± 2 The RHS of x 2 cannot be negative in order to yield x ∴ x = ± 2 − 2 + 8 = ± − 1 + 2 ∴ x = ± 2 − 2 + 8 = ± − 1 + 2 x = ± − 1 + 2 ≈ ± 0.6436 At x = 0.6436 , y = ± 0.6436 1 + 0.643 6 2 1 − 0.643 6 2 , y ≈ ± 0.4142. At x = − 0.6436 , y ≈ ∓ 0.4142 ∴ The stationary points are ( 0.6436 , 0.4142 ) , ( 0.6436 , − 0.4142 ) , ( − 0.6436 , 0.4142 ) , ( − 0.6436 , − 0.4142 ) . Calculating the x and y intercepts At x = 0 , y = 0 At y = 0 , x 1 − x 2 = 0 ⇒ x = 0 , x = − 1 , 1. ∴ The x intercept is ( − 1 , 0 ) , ( 0 , 0 ) & ( 1 , 0 ) and the y intercept is ( 0 , 0 ) . The graph is in the given below
