This question has to do with the Mathematical sub-topic in Geometry called Ellipse.
An ellipse is a circle that has been stretched in one direction so that it is no longer symmetrical but oval in shape.
Let us assume that the points on the ellipse is (cosθ, √3 sinθ)
Next let's minimize their distance squared from (-1, 0) taking a point on the ellipse as (cos θ- -1)² + (√3 sin θ)² =
cos²θ + 2cos θ + 1 + 3 sin² θ
(Substituting sin² θ + cos² θ = 1)
2 + 2 cos θ + 2 sin² θ
The derivative is given as
-2 sin θ + 4 sin θ cos θ =
-2 sin θ (2 cos θ - 1)
The above translates to 0 at sin θ = 0 or cos θ = 1/2, which results in 0°, 180°, 60°, 300°
Assuming we re-write the first derivative as - 2sin θ+ 2sin 2θ, the second derivative becomes:
-2 cos θ + 4 cos 2 θ
At 0°, 180°, 60°, 300°, we get 2, 6, -3, -3.
Therefore, our relative minima is given at 0°, 180°, while the relative maxima at 60°, 300°.
It is safe to declare, therefore that due to the nature of the function, the global minima and maxima will be among the four locations indicated above.
In addition to that,, 180° is the global minimum. This is because it is at the actual point, while 60° and 300° give the same value of the distance squared, thus both will each be maxima.
(cos θ, √3 sin θ) at 60°, 300° is (1/2, 3/2) and (1/2, -3/2)
Learn more about Point on Ellipse at:
https://brainly.com/question/4399570
See the definition of a circle here:
https://brainly.com/question/8952990
