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}}}.
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Question 1133939: Find the range of values of k, such that the straight line does not intersect the curve .
Found 2 solutions by greenestamps, JaydenX:
Answer by greenestamps(13198) (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.
[choose the half of the circle with y positive]
Now solve for the corresponding y value.
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.
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
Answer by JaydenX(2) (Show Source): You can put this solution on YOUR website!
Substitute into
Using the formula as it doesn’t intersect the curve.
Separate into possible cases,
and
Hence, the range of values of k,
,
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