Finding point of intersection where k is a non zero constant
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.
$\endgroup$1 Answer
$\begingroup$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.
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