y = kx - 2 ① , y² = 4x - x² ②

Substitute ① into ②,

(kx - 2)² = 4x - x²

k²x² - 4kx + 4 = 4x - x²

k²x² + x² - 4kx - 4x + 4 = 0

(k² + 1)x² - (4k + 4)x + 4 = 0

Interpreting 'meets the curve' as 'at only 1 point' i.e the line is tangent to the curve,

We need to solve for discriminant = 0
i.e 1 real root.

So b² - 4ac = 0

(-(4k + 4))² - 4(k² + 1)(4) = 0

16k² + 32k + 16 - 16k² - 16 = 0

32k = 0

k = 0

Set of values would be : ｛0｝

Interpreting 'meets the curve' as 'at 1 or more points' i.e the line is tangent to the curve or cuts the curve at 2 points

We need to solve for discriminant ≥ 0
i.e 1 or 2 real roots.

So b² - 4ac ≥ 0

(-(4k + 4))² - 4(k² + 1)(4) ≥ 0

16k² + 32k + 16 - 16k² - 16 ≥ 0

32k ≥ 0

k ≥ 0


Set of values would be : ｛k : k∈R, k ≥ 0｝

But it says sets of values so is it still possible for k >0?

Depends on how you interpret 'meets the curve'

Usually, it is taken to be at only 1 point.

So even if its 'set of values', it doesn't strictly have to be more than 1 value.

i.e just because the word used is plural, doesn't mean that its strictly more than 1 value.

Ok thank you