A normal to the hyperbola, $$4x^2 - 9y^2 = 36$$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the locus of P is:

We begin with the hyperbola $$4x^2-9y^2 = 36$$ and choose an arbitrary point on it, say $$P_1(x_1 ,\, y_1)$$. Because this point lies on the curve we already have

$$4x_1^2-9y_1^2 = 36 \;. \quad -(1)$$

To write the normal at this point, we first need the slope of the tangent. Differentiating $$4x^2-9y^2 = 36$$ implicitly with respect to $$x$$ we get

$$8x - 18y\,\dfrac{dy}{dx}=0,$$

so the derivative (slope of the tangent) is

$$\dfrac{dy}{dx} = \dfrac{8x}{18y} = \dfrac{4x}{9y}.$$

At the chosen point $$P_1(x_1,y_1)$$ the tangent slope is therefore

$$m_t = \dfrac{4x_1}{9y_1}.$$

The slope of the normal is the negative reciprocal of the tangent slope; hence

$$m_n = -\,\dfrac{1}{m_t}= -\,\dfrac{9y_1}{4x_1}.$$

The equation of the normal line through $$P_1(x_1,y_1)$$ then becomes

$$y-y_1 = -\,\dfrac{9y_1}{4x_1}\,(x-x_1). \quad -(2)$$

Next we find where this normal meets the coordinate axes.

x-intercept A (set $$y = 0$$ in (2)):

$$0 - y_1 = -\,\dfrac{9y_1}{4x_1}\,(x_A - x_1).$$

Dividing by $$-y_1$$ (assuming $$y_1 \neq 0$$) and simplifying,

$$1 = \dfrac{9}{4x_1}\,(x_A - x_1) \;\;\Longrightarrow\;\; x_A - x_1 = \dfrac{4x_1}{9} \;\;\Longrightarrow\;\; x_A = x_1 + \dfrac{4x_1}{9} = \dfrac{13x_1}{9}.$$

y-intercept B (set $$x = 0$$ in (2)):

$$y_B - y_1 = -\,\dfrac{9y_1}{4x_1}\,(0 - x_1) = \dfrac{9y_1}{4}.$$

Hence

$$y_B = y_1 + \dfrac{9y_1}{4} = \dfrac{13y_1}{4}.$$

Thus the intercepts are

$$A\left(\dfrac{13x_1}{9},\,0\right),\qquad B\left(0,\,\dfrac{13y_1}{4}\right).$$

The problem now forms the parallelogram $$OABP$$ with the origin $$O(0,0)$$. In a parallelogram, the fourth vertex is obtained by the vector sum of the position vectors of the adjacent vertices, so

$$\vec{OP} = \vec{OA} + \vec{OB}.$$

Consequently, the coordinates of $$P$$ are

$$P\;(x,y) = \left(\dfrac{13x_1}{9},\;\dfrac{13y_1}{4}\right).$$

We now express $$x_1$$ and $$y_1$$ in terms of $$x$$ and $$y$$:

$$x_1 = \dfrac{9x}{13},\qquad y_1 = \dfrac{4y}{13}.$$

Substituting these into the point condition (1) gives the required locus:

$$ 4\left(\dfrac{9x}{13}\right)^2 - 9\left(\dfrac{4y}{13}\right)^2 = 36. $$

Carrying out the squares,

$$ 4 \cdot \dfrac{81x^2}{169} \;-\; 9 \cdot \dfrac{16y^2}{169} = 36 \;\;\Longrightarrow\;\; \dfrac{324x^2}{169} \;-\; \dfrac{144y^2}{169} = 36. $$

Multiplying by $$169$$ clears the denominators:

$$324x^2 - 144y^2 = 36 \times 169 = 6084.$$

Finally we divide every term by $$36$$ to simplify:

$$9x^2 - 4y^2 = 169.$$

This equation represents a hyperbola and matches Option 3.

Hence, the correct answer is Option 3.