Respuesta :
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] .
There are four points where the tangent is horizontal: [tex](x_{1},y_{1}) = \left(+\frac{9\sqrt{3}}{8}, +\frac{9}{8}\right)[/tex], [tex](x_{2},y_{2}) = \left(+\frac{9\sqrt{3}}{8}, -\frac{9}{8}\right)[/tex], [tex](x_{3}, y_{3}) = \left(-\frac{9\sqrt{3}}{8}, +\frac{9}{8} \right)[/tex], [tex](x_{4}, y_{4}) = \left(-\frac{9\sqrt{3}}{8}, -\frac{9}{8} \right)[/tex].
The tangent line is horizontal when slope equals zero and the slope around the lemniscate can be determined by implicit differentiation:
[tex]8\cdot 2\cdot (x^{2}+y^{2}) \cdot (2\cdot x + 2\cdot y \cdot \dot y) = 81\cdot (2\cdot x - 2\cdot y \cdot \dot y)[/tex]
([tex]\dot y = 0[/tex])
[tex]32\cdot (x^{2}+y^{2}) \cdot x = 162\cdot x[/tex][tex]\left(\frac{162}{32}-2\cdot y^{2} \right)=\frac{81}{32}[/tex]
[tex]x^{2}+y^{2} = \frac{162}{32}[/tex] (1)
By (1):
[tex]x^{2} = \frac{162}{32} -y^{2}[/tex]
Then, the equation of the lemniscate is reduced into this form:
[tex]8\cdot \left(\frac{162}{32}\right)^{2} = 81\cdot \left(\frac{162}{32}-2\cdot y^{2} \right)[/tex]
[tex]\frac{162}{32}-2\cdot y^{2} =\frac{81}{32}[/tex]
[tex]2\cdot y^{2} = \frac{81}{32}[/tex]
[tex]y^{2} = \frac{81}{64}[/tex]
[tex]y = \pm \frac{9}{8}[/tex]
And the value of [tex]x[/tex] is:
[tex]x = \sqrt{\frac{162}{32}-\frac{81}{64} }[/tex]
[tex]x = \pm \frac{9\sqrt{3}}{8}[/tex]
There are four points where the tangent is horizontal: [tex](x_{1},y_{1}) = \left(+\frac{9\sqrt{3}}{8}, +\frac{9}{8}\right)[/tex], [tex](x_{2},y_{2}) = \left(+\frac{9\sqrt{3}}{8}, -\frac{9}{8}\right)[/tex], [tex](x_{3}, y_{3}) = \left(-\frac{9\sqrt{3}}{8}, +\frac{9}{8} \right)[/tex], [tex](x_{4}, y_{4}) = \left(-\frac{9\sqrt{3}}{8}, -\frac{9}{8} \right)[/tex].
We kindly invite to check this question on lemniscates: https://brainly.com/question/23025207
