There are at least three values of k for which, the point (k,1) lies on conic
ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0
State true or false and Justify
Substituting for the point (k,1) into the conic, we get
ak^2 + 2hk + b + 2gk + 2f + c = 0
ak^2 + (2h + 2g)k + (b + 2f + c) = 0
The above equation is a quadratic in k, and as such can have at most only two real solutions.
Therefore, the statement is false.