Find the points on the lemniscate where the tangent is horizontal. 8(x2 + y2)2 = 81(x2 − y2)

Answer:[tex](\dfrac{9\sqrt{3}}{8},\dfrac{9}{8}),(-\dfrac{9\sqrt{3}}{8},\dfrac{9}{8}),(\dfrac{9\sqrt{3}}{8},-\dfrac{9}{8}),(-\dfrac{9\sqrt{3}}{8},-\dfrac{9}{8})[/tex] Step-by-step explanation:The given equation of lemniscate is [tex]8(x^2+y^2)^2=81(x^2-y^2)[/tex] Differentiate with respect to x. [tex]16(x^2+y^2)(2x+2y\dfrac{dy}{dx})=81(2x-2y\dfrac{dy}{dx})[/tex] We know that [tex]\dfrac{dy}{dx}[/tex] is slope of the tangent and slope of a horizontal line is 0, So, [tex]\dfrac{dy}{dx}=0[/tex]. [tex]16(x^2+y^2)(2x+2y(0))=81(2x-2y(0))[/tex] [tex]32x(x^2+y^2)=162x[/tex] [tex]x^2+y^2=\dfrac{162x}{32x}[/tex] [tex]x^2+y^2=\dfrac{81}{16}[/tex] [tex]x^2=\dfrac{81}{16}-y^2[/tex] Substitute this value in the given equation. [tex]8(\dfrac{81}{16})^2=81(\dfrac{81}{16}-2y^2)[/tex] [tex](\dfrac{81}{16})(\dfrac{1}{2})=\dfrac{81}{16}-2y^2[/tex] [tex]y^2=-\dfrac{81}{32}+\dfrac{81}{16}[/tex] [tex]y^2=\dfrac{81}{64}[/tex] [tex]y=\pm \sqrt{\dfrac{81}{64}}[/tex] [tex]y=\pm \dfrac{9}{8}[/tex] Substitute the value of [tex]y^2[/tex] in the given equation. [tex]8(x^2+\dfrac{81}{64})^2=81(x^2\dfrac{81}{64})[/tex] On solving we get [tex]x=\pm \dfrac{9\sqrt{3}}{8}[/tex] Therefore, the required points are [tex](\dfrac{9\sqrt{3}}{8},\dfrac{9}{8}),(-\dfrac{9\sqrt{3}}{8},\dfrac{9}{8}),(\dfrac{9\sqrt{3}}{8},-\dfrac{9}{8}),(-\dfrac{9\sqrt{3}}{8},-\dfrac{9}{8})[/tex] .