Answer
The coordinates of the four points at which the tangent line is horizontal are $\left(\dfrac{\sqrt3}{\sqrt2},\dfrac{1}{\sqrt2}\right)$, $\left(-\dfrac{\sqrt3}{\sqrt2},\dfrac{1}{\sqrt2}\right)$, $\left(\dfrac{\sqrt3}{\sqrt2},-\dfrac{1}{\sqrt2}\right)$ and $\left(-\dfrac{\sqrt3}{\sqrt2},-\dfrac{1}{\sqrt2}\right)$.
Work Step by Step
Firstly differentiate $(x^2 + y^2)^2 = 4(x^2 − y^2)$ with respect to $x$.
We get, $\dfrac{d}{dx}\left((x^2 + y^2)^2\right)=4\dfrac{d}{dx}(x^2 − y^2)$
$\implies\dfrac{d}{dx}\left((x^2 + y^2)^2\right)=4\dfrac{d(x^2)}{dx}-4\dfrac{d(y^2)}{dx}$
Now use the chain rule as follows.
We get, $2(x^2+y^2)\dfrac{d}{dx}(x^2 + y^2)=8x-8y\dfrac{dy}{dx}$
$\implies 2(x^2+y^2)\left(2x+2y\dfrac{dy}{dx}\right)=8x-8y\dfrac{dy}{dx}$
Since, if the tangent line is horizontal, then $\dfrac{dy}{dx}=0$
Substitute $\dfrac{dy}{dx}=0$ in $2(x^2+y^2)\left(2x+2y\dfrac{dy}{dx}\right)=8x-8y\dfrac{dy}{dx}$.
We get, $2(x^2+y^2)(2x+2y(0))=8x-8y(0)$
Solve further as follows:
$4x(x^2+y^2)=8x$
$\implies x^2+y^2=2$
$\implies y^2=2-x^2$
Now substitute $y^2=2-x^2$ in $(x^2 + y^2)^2 = 4(x^2 − y^2)$.
We get, $(x^2 +2-x^2)^2 = 4(x^2 − 2+x^2)$
Simplify to find $x$.
We get, $(2)^2 = 4(2x^2 − 2)$ or $4=4(x^2-2)$
$\implies x^2=\dfrac{3}{2}$
$\implies x=\pm\dfrac{\sqrt3}{\sqrt2}$
Now subtitute $x=\dfrac{\sqrt3}{\sqrt2}$ and $x=-\dfrac{\sqrt3}{\sqrt2}$ one by one in $y^2=2-x^2$.
If we substitute $x=\dfrac{\sqrt3}{\sqrt2}$, then we get $y^2=2-\left(\dfrac{\sqrt3}{\sqrt2}\right)^2=2-\dfrac{3}{2}=\dfrac{1}{2}$.
$\implies y=\pm\dfrac{1}{\sqrt2}$
And if we substitute $x=-\dfrac{\sqrt3}{\sqrt2}$, then we get $y^2=2-\left(-\dfrac{\sqrt3}{\sqrt2}\right)^2=2-\dfrac{3}{2}=\dfrac{1}{2}$.
$\implies y=\pm\dfrac{1}{\sqrt2}$
So, the coordinates of the four points at which the tangent line is horizontal are $\left(\dfrac{\sqrt3}{\sqrt2},\dfrac{1}{\sqrt2}\right)$, $\left(-\dfrac{\sqrt3}{\sqrt2},\dfrac{1}{\sqrt2}\right)$, $\left(\dfrac{\sqrt3}{\sqrt2},-\dfrac{1}{\sqrt2}\right)$ and $\left(-\dfrac{\sqrt3}{\sqrt2},-\dfrac{1}{\sqrt2}\right)$.
