This problem requires us to find the locus of a point P, which is a vertex of a parallelogram formed by the origin O and the intercepts A and B of a normal to a hyperbola. We will systematically use concepts of the standard form of a hyperbola, its parametric representation, the equation of a normal, and properties of parallelograms.

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1. Standard Form of the Hyperbola and Parametric Point

The first crucial step is to convert the given hyperbola equation into its standard form, $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. This allows us to easily identify the semi-transverse axis ($a$) and semi-conjugate axis ($b$), which are essential for writing the parametric equations and the normal equation.

• Given Equation of Hyperbola: $4x^2 - 9y^2 = 36$

• Divide the entire equation by 36 to achieve the standard form: $\frac{4x^2}{36} - \frac{9y^2}{36} = \frac{36}{36}$ $\frac{x^2}{9} - \frac{y^2}{4} = 1$

• Identify the parameters $a^2$ and $b^2$: By comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we can deduce: $a^2 = 9 \implies a = 3$ $b^2 = 4 \implies b = 2$ (We take positive values for $a$ and $b$ as they represent lengths).

• Parametric Point on the Hyperbola: A general point $(x_1, y_1)$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ can be conveniently represented using a parameter $\theta$ as $(a \sec \theta, b \tan \theta)$. This choice of parametric form satisfies the hyperbola equation $\sec^2 \theta - \tan^2 \theta = 1$. Substituting our values for $a$ and $b$: $(x_1, y_1) = (3 \sec \theta, 2 \tan \theta)$ This point $(x_1, y_1)$ is the point on the hyperbola where the normal is drawn.

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2. Equation of the Normal to the Hyperbola

Next, we need to find the equation of the normal line to the hyperbola at our chosen parametric point $(x_1, y_1) = (3 \sec \theta, 2 \tan \theta)$. The general equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at a point $(x_1, y_1)$ is given by: $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$ This formula is derived from the fact that the slope of the tangent at $(x_1, y_1)$ is $m