Answer:
[tex]t_{1} = 0.422\pi \pm \pi\cdot i[/tex], [tex]\forall \,i \in \mathbb{N}_{O}[/tex]
[tex]t_{2} = 1.422\pi \pm \pi\cdot i[/tex], [tex]\forall \,i \in \mathbb{N}_{O}[/tex]
Step-by-step explanation:
In the case of parametric equations, the slope of the curve is equal to:
[tex]\frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
Where [tex]\frac{dx}{dt}[/tex] and [tex]\frac{dy}{dt}[/tex] are the first derivatives of [tex]x[/tex] and [tex]y[/tex] regarding [tex]t[/tex]. Let be [tex]x(t) =2\cdot \cos t[/tex] and [tex]y(t) = 8\cdot \sin t[/tex], their first derivatives are found:
[tex]\frac{dx}{dt} = -2\cdot \sin t[/tex] and [tex]\frac{dy}{dt} = 8\cdot \cos t[/tex]
Thus, equation for the slope is:
[tex]\frac{dy}{dx} = -\frac{8\cdot \cos t}{2\cdot \sin t}[/tex]
[tex]\frac{dy}{dx} = -\frac{4}{\tan t}[/tex]
If [tex]\frac{dy}{dx} = -1[/tex], then:
[tex]-1 = -\frac{4}{\tan t}[/tex]
[tex]\tan t = 4[/tex]
Tangent is positive at 1st quadrant and is a function with a periodicity of [tex]\pi[/tex], the set of solutions are:
[tex]t_{1} = 0.422\pi \pm \pi\cdot i[/tex], [tex]\forall \,i \in \mathbb{N}_{O}[/tex]
[tex]t_{2} = 1.422\pi \pm \pi\cdot i[/tex], [tex]\forall \,i \in \mathbb{N}_{O}[/tex]
