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Anonymous Tue 08 Oct 2019 19:15:37 No.723690 View View Reply Original Report

Quoted By:

Find the intersection between the following circumference and the coordinate axis.
(x-4)^2+y^2=16

Anonymous

Anonymous Tue 08 Oct 2019 19:18:03 No.723691 Report

Quoted By:

I seriously can't find the correct answer given by the professor.
It would be:

(0, 3 + √21), (0, 3 −√21), (2, 0) and (−6, 0).

Anonymous

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Anonymous Tue 08 Oct 2019 19:34:07 No.723695 Report

Quoted By: >>723700

> (2, 0) and (−6, 0)
but that's wrong
(2-4)^2 + 0^2 = 4, not 16
(-6-4)^2 + 0^2 = 100, not 16
The correct intersections with the x-axis are
(8, 0) and (0, 0)
Besides (0, 0) it doesn't intersect the y-axis in any other point.

Anonymous

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Anonymous Tue 08 Oct 2019 20:01:05 No.723700 Report

Quoted By:

>>723695
I fucking came over the same conclussion!
But when I'm asked, in another question, about calculating the intersection of the following elipse and the coordinate axis, what should I do?
Elipse -> 4(x-2)^2+9(y+1)^2=36

Anonymous

Anonymous Tue 08 Oct 2019 21:10:57 No.723727 Report

Quoted By:

>set x=0
4*4 + 9(y+1)^2 = 36
9(y+1)^2 = 20
(y+1)^2 = (20/9)
y+1 = +-(2/3)sqrt(5)
y = -1 +- (2/3)sqrt(5)
so
(0 , -1 - (2/3)sqrt(5))
(0 , -1 + (2/3)sqrt(5))

Now set y=0 and try to solve for x...

Anonymous

Anonymous Tue 08 Oct 2019 21:16:26 No.723729 Report

Quoted By:

4(x-2)^2 + 9 = 36
4(x-2)^2 = 27
(x-2)^2 = 27/4
x-2 = +- (3/2)sqrt(3)
x = 2 +- (3/2)sqrt(3)
so
(2 - (3/2)sqrt(3) , 0)
(2 + (3/2)sqrt(3), 0)

Don't forget to doublecheck btw.

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