The locus of a point $P(\alpha,\beta)$ moving under the condition that the line $y=\alpha x+\beta$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,is
(a) a hyperbola
(b) a parabola
(c) a circle
(d) an ellipse
What I tried...
The slope of the given line is $\alpha$, differentiating the equation of parabola and equating $\frac{dy}{dx}$ to $\alpha$, $$\frac{x}{a^2}=\frac{y}{b^2}\frac{dy}{dx}\implies \frac{x}{a^2}=\frac{y}{b^2}\cdot (\alpha)$$ From here, I have no clue how to proceed. Please help.
$y=\alpha x+\beta$ will be tangent to $x^2/a^2-y^2/b^2=1$ if it comapares to $y=mx+\sqrt{m^2a^2-b^2},$ we get $\alpha =m, \beta=\sqrt{m^2a^2-b^2}$. So, $\beta^2=a^2 \alpha^2-b^2 \implies a^2\alpha^2-\beta^2=b^2$. Finally the locus of $(\alpha, \beta)$ is $$\frac{x^2}{b^2/a^2}-\frac{y^2}{b^2}=1,$$ whiuch is a hyperbola.