Answer
$(±\frac{5\sqrt {3}}{4},±\frac{5}{4})$
Work Step by Step
a tangent to the lemniscate will be horizontal if $y'$ = $0$
$25x-4x(x^{2}+y^{2})$ = $0$
$x[25-4(x^{2}+y^{2})]$ = $0$
$x^{2}+y^{2}$ = $\frac{25}{4}$
Substituting $\frac{25}{4}$ in the equation of the lemniscate
$2(x^{2}+y^{2})^2$ = $25((x^{2}-y^{2}))$
$x^{2}-y^{2}$ = $\frac{25}{8}$
solve equation
$x^{2}+y^{2}$ = $\frac{25}{4}$ and
$x^{2}-y^{2}$ = $\frac{25}{8}$
Thus
$x^{2}$ = $\frac{75}{16}$
$y^{2}$ = $\frac{25}{16}$
so the $4$ points are $(±\frac{5\sqrt {3}}{4},±\frac{5}{4})$
