Find the sum of all possible values of the constant k such that the graph of the parametric equations x=2+4cos(s) and y=k-4sin(s) intersects the graph of the parametric equations x=1+cos(t) and y=-3+sin(t) at only one point.
For those of you who want to know, it's -12
To find the sum of all possible values of the constant k, we need to determine when the two parametric curves intersect at only one point.
First, let's set the x-components of the two curves equal to each other:
2 + 4cos(s) = 1 + cos(t)
Rearranging the equation, we have:
cos(s) = (1 + cos(t) - 2) / 4
cos(s) = (cos(t) - 1) / 4
Next, we equate the y-components of the two curves:
k - 4sin(s) = -3 + sin(t)
Rearranging this equation, we have:
sin(s) = (k + 3 - sin(t)) / 4
Now, since we want the two curves to intersect at only one point, we need the values of s and t to be the same for the x and y components. Therefore, we can equate the two values we just obtained for cos(s) and sin(s):
(cos(t) - 1) / 4 = (k + 3 - sin(t)) / 4
Simplifying this equation, we have:
cos(t) - 1 = k + 3 - sin(t)
Rearranging, we get:
cos(t) + sin(t) = 4 + k
Next, we can use the trigonometric identity cos^2(t) + sin^2(t) = 1 to replace sin(t) in the equation:
cos(t) + sqrt(1 - cos^2(t)) = 4 + k
Simplifying further, we have:
cos(t) + sqrt(1 - cos^2(t)) - 4 = k
Now, we need to determine when the equation above only has one solution for k. This occurs when the discriminant of the quadratic inside the square root is equal to zero. So we set it to zero and solve for cos(t):
1 - cos^2(t) = 0
cos^2(t) = 1
Taking the square root of both sides, we have:
cos(t) = ±1
This means that t can take on the values of 0 and π. Hence, substituting these values back into the equation for k, we get:
k = cos(0) + sqrt(1 - cos^2(0)) - 4
k = cos(π) + sqrt(1 - cos^2(π)) - 4
Evaluating these expressions, we find:
k = 2 + 0 - 4 = -2
k = -1 + 0 - 4 = -5
Finally, to find the sum of all possible values of k, we add these two values together:
Sum of all possible values of k = -2 + (-5) = -7
(x-2)^2/4 + (y-k)^2/16 = 1
(x-1)^2 + (y+3)^2 = 1
We have an ellipse with center at (2,k) and radii 2 and 4
and a circle with center (1,-3) with radius 1
They are tangent internally at (0,-3) if k = -3
If they are tangent externally, then there will be two values at k = -3±c for some value of c.
So, regardless of just what c is, I get -9 as the sum of all possible values of k.