SOLUTION: Find the values of k for which the line {{{y + kx = 8}}} is a tangent to the curve {{{(x)^(2) + 4y = 20}}}. *PLEASE ANSWER AS SOON AS POSSIBLE BRO :)

Algebra ->  Coordinate-system -> SOLUTION: Find the values of k for which the line {{{y + kx = 8}}} is a tangent to the curve {{{(x)^(2) + 4y = 20}}}. *PLEASE ANSWER AS SOON AS POSSIBLE BRO :)       Log On


   

Question 475425: Find the values of k for which the line y+%2B+kx+=+8 is a tangent to the curve %28x%29%5E%282%29+%2B+4y+=+20.

*PLEASE ANSWER AS SOON AS POSSIBLE BRO :)
Found 2 solutions by ewatrrr, richard1234:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

Find the values of k for which the line y+%2B+kx+=+8 is a tangent to the curve %28x%29%5E%282%29+%2B+4y+=+20.

%28x%29%5E%282%29+%2B+4y+=+20 OR y = -.25x^2 +5 Parabola V(0,5)

 y = -kx + 8   Line would be tangent to the Parabola at Pt(3,3)

 3 = -k*3 + 8   k = 5/3

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
We could use calculus or a graph (actually, the graph still helps as it gives a good visual), but the easiest way to do it is to treat them as functions of y and set them equal.


Since the quadratic is differentiable everywhere, we can say that a tangency point occurs when the equation has only one solution for x. We turn this into a quadratic in terms of x

, set the discriminant equal to zero.