SOLUTION: Find the range of values of k, such that the straight line {{{y=3x+k}}} does not intersect the curve {{{x^2+y^2-8=0}}}.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Find the range of values of k, such that the straight line {{{y=3x+k}}} does not intersect the curve {{{x^2+y^2-8=0}}}.      Log On


   

Question 1133939: Find the range of values of k, such that the straight line y=3x%2Bk does not intersect the curve x%5E2%2By%5E2-8=0.
Found 2 solutions by greenestamps, JaydenX:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The lines of the form y=3x+k have slope 3.

The curve x^2+y^2-8=0 is a circle with center at the origin and radius 2*sqrt(2); there will be two points on the circle where the slope of the curve -- and therefore the slope of a tangent to the curve -- is 3.

(1) Find the points where the slope of the curve is 3.
(2) Find the equations of the lines tangent to the curve at those points. The values of k for which the line y=3x+k DOES intersect the circle are the values between the y-intercepts of those two lines.

Note: Because the origin is the center of the circle, the two solutions will be symmetrical; we only need to do the complete solution process for one of them.

If we choose the half of the circle with y positive, a line with slope 3 will have a positive y-intercept; and the x value will be negative.

x%5E2%2By%5E2-8+=+0
y+=+sqrt%288-x%5E2%29+=+%288-x%5E2%29%5E%281%2F2%29 [choose the half of the circle with y positive]
dy%2Fdx+=+%281%2F2%29%288-x%5E2%29%5E%28-1%2F2%29%28-2x%29+=+-x%2F%28sqrt%288-x%5E2%29%29

-x%2Fsqrt%288-x%5E2%29+=+3
-x%2F3+=+sqrt%288-x%5E2%29
x%5E2%2F9+=+8-x%5E2
%2810%2F9%29x%5E2+=+8
x%5E2+=+72%2F10+=+36%2F5
x+=+-6%2Fsqrt%285%29

Now solve for the corresponding y value.

%28-6%2Fsqrt%285%29%29%5E2%2By%5E2+=+8
36%2F5%2By%5E2+=+8+=+40%2F5
y%5E2+=+4%2F5
y+=+2%2Fsqrt%285%29

The point on the half of the circle where y is positive is (-6/sqrt(5),2/sqrt(5))

Now find the equation of the line with slope 3 through that point.

2%2Fsqrt%285%29+=+3%28-6%2Fsqrt%285%29%29%2Bk
k+=+20%2Fsqrt%285%29+=+4%2Asqrt%285%29

By symmetry, the two values of k for which the line with 3x+k just touches the circle are 4*sqrt(5) and -4*sqrt(5). So....

ANSWER: The line y=3x+k does not intersect the curve x^2+y^2-8=0 for value of k for which abs%28k%29%3E4%2Asqrt%285%29

Answer by JaydenX(2) About Me  (Show Source):
You can put this solution on YOUR website!
Substitute y=3x%2Bk into x%5E2%2By%5E2-8=0
x%5E2%2B%283x%2Bk%29%5E2-8=0
x%5E2%2B9x%5E2%2B6kx%2Bk%5E2-8=0
10x%5E2%2B6kx-8=0
Using the formula b%5E2-4ac%3C0 as it doesn’t intersect the curve.
%286k%29%5E2-4%2810%29%28k%5E2-8%29%3C0
36k%5E2-40k%5E2%2B320%3C0
-4k%5E2%2B320%3C0
4k%5E2%3E320
Separate into possible cases,
k%3Esqrt%2880%29 and -k%3Esqrt%2880%29
Hence, the range of values of k,
k%3Esqrt%2880%29 , k%3C-sqrt%2880%29