Question

Find the points on the ellipse 3x2 + y2 = 3 that are farthest away from the point (1, 0).

Answer 1

This question has to do with the Mathematical sub-topic in Geometry called Ellipse.

What is an Ellipse in Maths?

An ellipse is a circle that has been stretched in one direction so that it is no longer symmetrical but oval in shape.

How does one find the points on the ellipse?

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:
brainly.com/question/4399570

See the definition of a circle here:
brainly.com/question/8952990