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?

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

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Answer 1 · 2018-02-28T12:56:49

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 $-10$