$x+2y-1=0; x^2-2y^2=n$, line should be tangent to the hyperbole, solve for n

Question

could you help me, please?

$x+2y-1=0; x^2-2y^2=n$;

Solve for n. What I did (wrongly):

$x+2y=1; (x^2-2y^2)/n=1; x+2y=(x^2-2y^2)/n; n(x+2y)=(x-2y)(x+2y); n=x-2y$

What does this relationship mean?

The solution should be $n=-1$

Answer 1

A line is tangent to a hyperbola if and only if their intersection consists of a single point. So, solve the system$$\left\{\begin{array}{l}x+2y-1=0\\x^2-2y^2=n.\end{array}\right.$$You can easily see that it has two solutions if $n>-1$, one solution if $n=-1$ and no solutions otherwise. That's why that answer is $n=-1$.