Question

Find the points on x^2+3(y-2)^2 = 27 which are closest to and farthest from the origin

Answer 1

$(-4.89, 3)$ and $(4.89, 3)$ are the farthest from the origin and $(0, -1)$ is the closest from the origin.

Let be a curve of the form $x^{2}+3\cdot (y-2)^{2} = 27$. The distance with respect to origin is found by using the following Pythagorean identity:

$r = x^{2}+y^{2}$ (1)

Where $r$ is the square distance function.

We can modify (1) as following:

$r = [27-3\cdot (y-2)^{2}]+y^{2}$

$r = 27-3\cdot (y^{2}-4\cdot y +4)+y^{2}$

$r = 27-3\cdot y^{2}+12\cdot y -12 + y^{2}$

$r = -2\cdot y^{2}+12\cdot y+27$ (1b)

Now we apply the First and second derivative tests to determine the minimum and maximum distances from the origin:

First derivative test

$-4\cdot y +12 = 0$

$y = 3$

Second derivative test

$r'' = -4$

The y-component represents a maximum.

Now we graph the function with the ressource of a graphing tool, we find the following points:

Farthest points: $(-4.89, 3)$, $(4.89, 3)$.

Closest points: $(0, -1)$.

$(-4.89, 3)$ and $(4.89, 3)$ are the farthest from the origin and $(0, -1)$ is the closest from the origin.

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