Answer to Question #183259 in Calculus for Nahiyan Anwar Orko

Question

Find all the points on the graph of (x^2 + y^2)^2 = x^2−y^2 that have horizontal tangent line.

Answer 1

Horizontal tangent at the point $x=x_0$ requires $y'(x=x_0)=0$.

Let's apply operator $\frac{d}{dx}$ to our expression, $y$ being function $y(x)$:

$x^4 +2x^2y^2+y^2=x^2-y^2 \quad \bigg | \frac{d}{dx}$

$4x^3+4xy^2+4x^2yy'+4y^3y'=2x-2yy'$

Substitute $x=x_0$ into this expression, remembering $y'(x=x_0)=0$ :

$4x_0^3+4x_0y(x_0)=2x_0$

Let $y(x_0)=y_0$. We can also substitute $x=x_0$ into our initial expression. Then we get the system of two equations and $y_0$ and $x_0$ can be found.

$\begin{cases} 4x_0^3+4x_0y_0^2=2x_0\\ (x_0^2+y_0^2)^2=x_0^2-y_0^2 \end{cases}\\ \begin{cases} x_0^2+y_0^2=\frac{1}{2}\\ (x_0^2+y_0^2)^2=x_0^2-y_0^2 \end{cases}\\ \begin{cases} x_0^2=\frac{1}{2}-y_0^2\\ \frac{1}{4}=\frac{1}{2}-2y_0^2 \end{cases}\\ \begin{cases} x_0^2=\frac{3}{8}\\ y_0^2=\frac{1}{8} \end{cases}$

So we got four points where tanget is horizontal: $(\frac{1}{\sqrt{8}},\sqrt{\frac{3}{8}}), (-\frac{1}{\sqrt{8}},\sqrt{\frac{3}{8}}), (\frac{1}{\sqrt{8}},-\sqrt{\frac{3}{8}}), (-\frac{1}{\sqrt{8}},-\sqrt{\frac{3}{8}})$