cant do curve question (1 Viewer)

[ek102]

The equation of a parabola is of the form y=kx^2. If the line 8x-y-4=0 is a tangent to the parabola, find the value of k.
thanks.

 

[pikachu975]

The line is a tangent when the gradient of the line is equal to the derivative.

Gradient of the line:
y = 8x-4
m = 8

y' = 2kx

So 2kx = 8
k = 4/x

Now do simultaneous equations for y1 = y2
kx^2 = 8x-4
k = (8x-4)/x^2
sub in k=4/x
4/x = (8x-4)/x^2
4x = (8x-4)
4x = 4
x = 1

Sub x = 1 into k
k = 4/1
k = 4

 

[KingOfActing]

 

[Drongoski]

Or

.: y = kx² and y = 8x - 4

Where these 2 curves meet:

kx² = 8x-4 ==> the quadratic eqn: kx² - 8x + 4 = 0

Since one is tangent to the parabola, this eqn has one repeated root.

.: its discriminant = 8² - 4 * k * 4 = 0 ==> k = 4

 

[frog1944]

I'd do it the way Drongoski did it