Question

1. Given that the straight line y=c-3x does not intersect the curve xy=3, find the range of values for c.

2. Find the range of values for c for which the line y=cx+6 does not meet the curve 2x^2-xy=3.

3. Find the range of values for k for which 8y=x+2k is a tangent to the curve 2y^2=x+k

Answer 1

The slope of line cannot be that of curve or greater.

slope line=-3

slope curve>slope line

-3<-3/x^2
x^2>1
x>1
check x=10 slope line=-3
slope curve=-3/100=-.03
check x=1/2 slope curve=-12
do the others the same way

Answer 2

number one is supposed to be -6<c<6
number two is supposed to be c>5
and number three is supposed to be k=8

Answer 3

the slope of xy=3 is y' = -3/x^2
So, the slope is -3 at x=1

The tangent to the curve at x=1 is

y-3 = -3(x-1)
y = -3x+6
y = 6-3x

so, for y = c-3x, if c < 6, the line falls below the curve. Similarly, for x<0, the line lies above the other branch of the curve if c > -6.

So, -6 < c < 6

Follow this logic for the other parts

Answer 4

Bruh i don't understand yallz explanation