Find the value(s) or range of values of k for which
(a) the line y = kx - 5 is tangent to the curve x^2 = 2y + 1.
(b) the line x + 3y = k - 1 intersect the curve y^2 = 2x + 5.
+130071 (a) the line y = kx - 5 is tangent to the curve x^2 = 2y + 1
Using implicit differentiation, the slope at any point on the quadratic function (curve) can be found as
2y = x^2 - 1
2y' = 2x
y' = x
Let the point where the line is tangent to the curve be (x , x^2/2 - 1/2)
The line will intercept the point (0,-5)
So we need to solve the slope equation
( x^2/2 - 1/2 - - 5) / ( x - 0) = x
(x^2/2 - 1/2 + 5) / x = x multiply through by x
x^2/2 + 9/2 = x^2 multiply through by 2
x^2 + 9 = 2x^2 rearrange as
x^2 - 9 = 0
x^2 = 9 take both roots
x = 3 or x = -3 = the two values for k
The lines are
y = 3x - 5 and y = -3x -5
See here : https://www.desmos.com/calculator/npleonadqb
