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

The given line is $$y=\alpha x+\beta$$ and the given conic is the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$.

Step 1 Substitute the line in the hyperbola.
Putting $$y=\alpha x+\beta$$ in the hyperbola, we get

$$\frac{x^{2}}{a^{2}}-\frac{(\alpha x+\beta)^{2}}{b^{2}}=1.$$

Multiply by $$a^{2}b^{2}$$ to clear denominators:

$$b^{2}x^{2}-a^{2}(\alpha x+\beta)^{2}-a^{2}b^{2}=0.$$

Step 2 Write the resulting quadratic in $$x$$.
Expand and collect terms of powers of $$x$$:

$$\bigl(b^{2}-a^{2}\alpha^{2}\bigr)x^{2}-2a^{2}\alpha\beta\,x-a^{2}\bigl(\beta^{2}+b^{2}\bigr)=0.$$

This is a quadratic of the form $$A x^{2}+B x+C=0$$ with
$$A=b^{2}-a^{2}\alpha^{2},\quad B=-2a^{2}\alpha\beta,\quad C=-a^{2}\bigl(\beta^{2}+b^{2}\bigr).$$

Step 3 Impose the condition for tangency.
A line is tangent to a conic when this quadratic has equal roots, i.e. its discriminant is zero:

$$B^{2}-4AC=0.$$

Substitute $$A,B,C$$:

$$\bigl(-2a^{2}\alpha\beta\bigr)^{2}-4\bigl(b^{2}-a^{2}\alpha^{2}\bigr)\!\bigl(-a^{2}(\beta^{2}+b^{2})\bigr)=0.$$

Simplify (divide by the common factor $$4$$ first):

$$a^{4}\alpha^{2}\beta^{2}+a^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)\bigl(\beta^{2}+b^{2}\bigr)=0.$$

Expand the second product:

$$a^{4}\alpha^{2}\beta^{2}+a^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)\beta^{2}+a^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)b^{2}=0.$$

Group the terms containing $$\beta^{2}$$:

$$\beta^{2}\,a^{2}\Bigl(a^{2}\alpha^{2}+b^{2}-a^{2}\alpha^{2}\Bigr)+a^{2}b^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)=0.$$

Inside the parenthesis $$\bigl(a^{2}\alpha^{2}+b^{2}-a^{2}\alpha^{2}\bigr)=b^{2}$$, therefore

$$a^{2}b^{2}\beta^{2}+a^{2}b^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)=0.$$

Divide both sides by the non-zero factor $$a^{2}b^{2}$$:

$$\beta^{2}+b^{2}-a^{2}\alpha^{2}=0.$$

Step 4 Obtain the locus equation.
Re-arrange:

$$a^{2}\alpha^{2}-\beta^{2}=b^{2}.$$

Divide by $$b^{2}$$ to see the standard form:

$$\frac{a^{2}\alpha^{2}}{b^{2}}-\frac{\beta^{2}}{b^{2}}=1$$
or equivalently
$$\frac{\alpha^{2}}{\,b^{2}/a^{2}\,}-\frac{\beta^{2}}{b^{2}}=1.$$

Step 5 Identify the curve.
This equation has a positive $$\alpha^{2}$$ term and a negative $$\beta^{2}$$ term, matching the general form $$\frac{X^{2}}{A^{2}}-\frac{Y^{2}}{B^{2}}=1$$ of a hyperbola. Hence the locus of $$P(\alpha,\beta)$$ is a hyperbola.

Option D which is: a hyperbola