# Refer to image. Thanks

### 1 Answer | Add Yours

Given parabola

`y=x^2+2` (i)

and line L posses through (0,1) , which is parallel to x-axis.

Thus equation of line L is

`y=1` (ii)

Parabola does not intersect line L ,so there are no points common in (i) and (ii).

In other words

`x^2+2=1` has no real solution.

### Hide Replies ▲

Please add this to answer to complete the answer

`x^2+2=k`

`x^2=k-2`

`x=+-sqrt(k-2)`

Thus x has two distinct and real values if k -2 >0

i.e.

k > 2