Question

The line (k-2)y = 3x meets the curve xy= 1 -x at two distinct points, Find the set of values of k. State also the values of k if the line is a tangent to the curve. How to find it?

The answers from the book is
k<-10 or k>2

Answer 1

the equation of the line can be rewritten as
`((k-2)y)/3 = x`

Substituting the value of x in the equation of the curve,

`(((k-2)y)/3)y = 1-((k-2)y)/3`

let `k-2 = a`

`(y^2a)/3 = (3-ya)/3`

`y^2a+ya-3=0`

Since the line intersects at two different points, the discriminant of the above equation must be greater than zero.

`D = a^2-4(-3)(a)>0`

`a[a+12]>0`

The range of `a` comes out to be,
`a in (-oo,-12)uu(0,oo)`

therefore,

`(k-2) in (-oo,-12)uu(2,oo)`

Adding 2 to both sides,

`k in (-oo,-10),(2,oo)`

If the line has to be a tangent, the discriminant must be zero, because it only touches the curve at one point,

`a[a+12] = 0`

`(k-2)[k-2+12]=0`

So, the values of `k` are `2` and `-10`