I am struggling to solve this question:
The curve $C$ has the equation $$ k x^2 - xy + (k+1)x=1. $$
The line $l$ has the equation $$ -(k/2)x + y = 1. $$
Here $k$ is a non-zero constant such that $l$ and $C$ only intersect at one point.
Find the coordinates of intersection of the line and the curve.
Any help would be appreciated.
From the equation of the line, we can solve for $y$ to obtain $$ y = (k/2)x + 1. $$ Substitute this value for $y$ into the equation of the curve to obtain $$ k x^2 - x \left( (k/2)x + 1 \right) + (k+1) x = 1, $$ which implies that $$ (k/2) x^2 + k x - 1 = 0, $$ which is a quadratic equation in $x$, and this equation has a unique solution if and only if the discriminant vanishes, that is, if and only if $$ k^2 + 4 (k/2) = 0, $$ which is the case if and only if $$ k = 0 \qquad \mbox{ or } \qquad k = -2. $$
Now we have the following two cases:
Case 1. When $k = 0$, the line has the equation $$ y = 1, $$ whereas the curve has the equation $$ -xy + x = 1. $$ Putting $y = 1$ into the equation $$ -xy + x = 1, $$ we obtain $$ -x + x = 1, $$ or $0 = 1$, which is impossible. So $k$ cannot be $0$.
Case 2. When $k = -2$, the line has the equation $$ -x + y = 1, \tag{1} $$ whereas the curve has the equation $$ -2x^2 - xy - x = 1, $$ which is equivalent to $$ 2x^2 + xy + x = -1. \tag{2} $$ Hope you know how to solve (1) and (2) simultaneously.