1. Fundamental Concept: Condition for Tangency to a Hyperbola

To determine the locus of point $P$, the most critical piece of knowledge required is the condition under which a straight line is tangent to a hyperbola.

For a standard hyperbola with the equation: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ A straight line given by the equation $y = mx + c$ will be tangent to this hyperbola if and only if the following condition is satisfied: $c^2 = a^2m^2 - b^2$

Optional Derivation for Deeper Understanding: This condition is derived by substituting the line equation $y = mx+c$ into the hyperbola equation. This process will yield a quadratic equation in $x$. For the line to be tangent, it must intersect the hyperbola at exactly one point. This implies that the quadratic equation must have exactly one real root, which means its discriminant ($\Delta$) must be equal to zero.

Let's perform the substitution: $\frac{x^2}{a^2} - \frac{(mx+c)^2}{b^2} = 1$ Multiply by $a^2b^2$ to clear denominators: $b^2x^2 - a^2(mx+c)^2 = a^2b^2$ Expand the squared term: $b^2x^2 - a^2(m^2x^2 + 2mcx + c^2) = a^2b^2$ Rearrange into the standard quadratic form $Ax^2 + Bx + C = 0$: $(b^2 - a^2m^2)x^2 - (2a^2mc)x - (a^2c^2 + a^2b^2) = 0$ For tangency, the discriminant must be zero: $\Delta = B^2 - 4AC = 0$. Here, $A = (b^2 - a^2m^2)$, $B = -2a^2mc$, and $C = -(a^2c^2 + a^2b^2)$. $(-2a^2mc)^2 - 4(b^2 - a^2m^2)(-(a^2c^2 + a^2b^2)) = 0$ $4a^4m^2c^2 + 4(b^2 - a^2m^2)(a^2c^2 + a^2b^2) = 0$ Divide the entire equation by $4a^2$ (assuming $a \neq 0$): $a^2m^2c^2 + (b^2 - a^2m^2)(c^2 + b^2) = 0$ Expand the product: $a^2m^2c^2 + b^2c^2 + b^4 - a^2m^2c^2 - a^2m^2b^2 = 0$ Notice that the $a^2m^2c^2$ terms cancel out: $b^2c^2 + b^4 - a^2m^2b^2 = 0$ Divide by $b^2$ (assuming $b \neq 0$): $c^2 + b^2 - a^2m^2 = 0$ Rearranging this equation gives us the tangency condition: $c^2 = a^2m^2 - b^2$ This confirms the validity of the condition.

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2. Analyzing the Problem Statement and Defining the Goal

We are given the following information:

• A point $P(\alpha, \beta)$ whose locus we need to determine. The locus is the path traced by $P$ as it moves according to the given conditions.
• A straight line defined by the equation $y = \alpha x + \beta$.
• The crucial condition that this line is tangent to the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Our objective is to find an algebraic equation that relates $\alpha$ and $\beta$. This equation will represent the locus of point $P$. Once we have this equation, we will identify the type of conic section it represents.

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3. Step-by-Step Derivation of the Locus

Let's systematically apply the tangency condition to the given problem.

• Step 1: Identify the parameters ($m$ and $c$) of the given tangent line. The general form of a straight line for the tangency condition is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. The line given in the problem is $y = \alpha x + \beta$. Why this step is taken: To use the standard tangency condition ($c^2 = a^2m^2 - b^2$), we must first identify the slope and y-intercept of the given line in terms of the coordinates of point $P(\alpha, \beta)$. This step effectively maps the problem's specific parameters to the general formula's variables. By comparing $y = \alpha x + \beta$ with $y = mx + c$, we can directly identify: $m = \alpha$ $c = \beta$

• Step 2: Apply the tangency condition using the identified parameters. Now that we have established $m = \alpha$ and $c = \beta$, we substitute these values into the tangency condition for the hyperbola: $c^2 = a^2m^2 - b^2$. Why this step is taken: The problem explicitly states that the line $y = \alpha x + \beta$ is tangent to the hyperbola. Therefore, its slope and y-intercept must satisfy the tangency condition. This step translates the geometric condition (tangency) into an algebraic relationship involving $\alpha$ and $\beta$, which are the coordinates of our moving point $P$. Substituting $m = \alpha$ and $c = \beta$ into the condition, we get: $(\beta)^2 = a^2(\alpha)^2 - b^2$ This simplifies to: $\beta^2 = a^2\alpha^2 - b^2$

• Step 3: Rearrange the equation to represent the locus. The equation $\beta^2 = a^2\alpha^2 - b^2$ is an algebraic relationship between the coordinates $(\alpha, \beta)$ of the point $P$. This equation is the locus. To express the locus in its conventional form, we replace $\alpha$ with $x$ and $\beta$ with $y$, as $(x, y)$ are typically used to denote a general point on a curve. Why this step is taken: The locus of a point is the equation that its coordinates satisfy as it moves according to the given conditions. By replacing $\alpha$ with $x$ and $\beta$ with $y$, we obtain the standard form of the equation representing the curve traced by $P$, making it easier to identify. Rearranging the equation to a more recognizable form for conic sections: $a^2\alpha^2 - \beta^2 = b^2$ Replacing $\alpha$ with $x$ and $\beta$ with $y$ to represent the locus: $a^2x^2 - y^2 = b^2$

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4. Identifying the Type of Curve

The equation of the locus is $a^2x^2 - y^2 = b^2$. To identify this curve, we can divide the entire equation by $b^2$ (assuming $b \neq 0$): $\frac{a^2x^2}{b^2} - \frac{y^2}{b^2} = \frac{b^2}{b^2}$ $\frac{x^2}{(b/a)^2} - \frac{y^2}{b^2} = 1$ This equation is of the general form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, where $A = b/a$ and $B = b$. This is the standard form of a hyperbola centered at the origin, with its transverse axis along the x-axis.

Common Mistake to Avoid: It is crucial not to confuse this form with an ellipse. An ellipse would have a positive sign between the $x^2$ and $y^2$ terms (e.g., $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$). The defining characteristic of a hyperbola is the negative sign between the squared terms when they are on the same side of the equation.

Therefore, the locus of point $P(\alpha, \beta)$ is a hyperbola.

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5. Conclusion and Key Takeaway

The locus of a point $P(\alpha, \beta)$ for which the line $y = \alpha x + \beta$ is tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by the equation $a^2x^2 - y^2 = b^2$. This equation represents a hyperbola.

The correct option is (D).

Key Takeaway for Locus Problems Involving Tangency: When asked to find the locus of a point whose coordinates are involved in the equation of a line tangent to a conic section:

1. Recall the tangency condition: Memorize the conditions for tangency for standard conic sections (parabola, ellipse, hyperbola, circle).
2. Map parameters: Identify the slope ($m$) and y-intercept ($c$) of the given tangent line in terms of the coordinates of the moving point.
3. Apply the condition: Substitute these identified parameters into the tangency condition.
4. Formulate the locus: The resulting equation, after replacing the moving point's coordinates (e.g., $\alpha, \beta$) with standard variables ($x, y$), will be the equation of the locus.
5. Identify the curve: Recognize the standard form of the resulting equation to identify the type of curve.