Tangents are drawn to the hyperbola $$4x^2 - y^2 = 36$$ at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of $$\triangle PTQ$$ is:

We are given the hyperbola $$4x^{2}-y^{2}=36$$. Dividing each term by $$36$$ we write it in standard form

$$\frac{x^{2}}{9}-\frac{y^{2}}{36}=1.$$

Hence $$a^{2}=9$$ and $$b^{2}=36$$ for the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1.$$

For this hyperbola the slope-form (also called the $$m$$-form) of a tangent is

$$y = mx \;\pm\; \sqrt{a^{2}m^{2}-b^{2}}.$$

We know that the two tangents meet at the fixed point $$T(0,3)$$. If the slope of a tangent is $$m$$, its equation through the point $$T(0,3)$$ is obtained from the two-point form:

$$y-3 = m(x-0) \;\Longrightarrow\; y = mx + 3.$$

This very same line must also fit the slope-form, so we equate the constant terms. Comparing

$$y = mx + 3 \quad\text{with}\quad y = mx \;\pm\; \sqrt{a^{2}m^{2}-b^{2}},$$

we require

$$\sqrt{a^{2}m^{2}-b^{2}} = 3 \quad\text{or}\quad \sqrt{a^{2}m^{2}-b^{2}} = -3.$$

Because a square-root is non-negative we take the positive sign and square both sides:

$$a^{2}m^{2}-b^{2} = 9.$$

Substituting $$a^{2}=9$$ and $$b^{2}=36$$ we obtain

$$9m^{2}-36 = 9,$$

$$9m^{2} = 45,$$

$$m^{2} = 5,$$

so

$$m = \sqrt{5}\quad\text{or}\quad m = -\sqrt{5}.$$

Thus the two tangents are

$$\text{(i)}\; y = \sqrt{5}\,x + 3,$$

$$\text{(ii)}\; y = -\sqrt{5}\,x + 3.$$

To find the points of contact $$P$$ and $$Q$$ we intersect each tangent with the hyperbola, and because tangency means a double root, we can proceed directly.

Tangent with slope $$m=\sqrt{5}$$

Substitute $$y = \sqrt{5}\,x + 3$$ into $$4x^{2}-y^{2}=36$$:

$$4x^{2} - \left(\sqrt{5}\,x + 3\right)^{2} = 36.$$

Expand the square:

$$4x^{2} - \left(5x^{2} + 6\sqrt{5}\,x + 9\right) = 36.$$

Simplify term by term:

$$4x^{2} - 5x^{2} - 6\sqrt{5}\,x - 9 = 36,$$

$$-x^{2} - 6\sqrt{5}\,x - 9 = 36,$$

$$-x^{2} - 6\sqrt{5}\,x - 45 = 0.$$

Multiply through by $$-1$$:

$$x^{2} + 6\sqrt{5}\,x + 45 = 0.$$

The discriminant is

$$\Delta = (6\sqrt{5})^{2} - 4\cdot 1 \cdot 45 = 180 - 180 = 0,$$

confirming tangency. The repeated root gives the $$x$$-coordinate of the point of contact:

$$x = -\frac{6\sqrt{5}}{2} = -3\sqrt{5}.$$

Put this value into the tangent equation to get $$y$$:

$$y = \sqrt{5}\,(-3\sqrt{5}) + 3 = -15 + 3 = -12.$$

So

$$P\bigl(-3\sqrt{5},\,-12\bigr).$$

Tangent with slope $$m=-\sqrt{5}$$

Substitute $$y = -\sqrt{5}\,x + 3$$ into $$4x^{2}-y^{2}=36$$:

$$4x^{2} - \left(-\sqrt{5}\,x + 3\right)^{2} = 36.$$

Expand the square:

$$4x^{2} - \left(5x^{2} - 6\sqrt{5}\,x + 9\right) = 36.$$

Simplify term by term:

$$4x^{2} - 5x^{2} + 6\sqrt{5}\,x - 9 = 36,$$

$$-x^{2} + 6\sqrt{5}\,x - 9 = 36,$$

$$-x^{2} + 6\sqrt{5}\,x - 45 = 0.$$

Multiply through by $$-1$$:

$$x^{2} - 6\sqrt{5}\,x + 45 = 0.$$

The discriminant is again zero:

$$\Delta = (-6\sqrt{5})^{2} - 4\cdot 1 \cdot 45 = 180-180=0,$$

so the double root is

$$x = \frac{6\sqrt{5}}{2} = 3\sqrt{5}.$$

The $$y$$-coordinate is

$$y = -\sqrt{5}\,(3\sqrt{5}) + 3 = -15 + 3 = -12.$$

Therefore

$$Q\bigl(3\sqrt{5},\,-12\bigr).$$

We now possess the three vertices of the triangle:

$$P(-3\sqrt{5},-12), \quad Q(3\sqrt{5},-12), \quad T(0,3).$$

The line $$PQ$$ is horizontal because both points share the same $$y$$-coordinate $$-12$$, so

$$\text{base } PQ = |\,3\sqrt{5} - (-3\sqrt{5})\,| = 6\sqrt{5}.$$

The altitude from $$T$$ to the base is the vertical distance between $$y=3$$ and $$y=-12$$:

$$\text{height } = 3 - (-12) = 15.$$

Using the formula for the area of a triangle, $$\text{Area} = \dfrac{1}{2}\times\text{base}\times\text{height},$$ we get

$$\text{Area} = \frac{1}{2}\times (6\sqrt{5}) \times 15 = 3\sqrt{5}\times 15 = 45\sqrt{5}.$$

Hence, the correct answer is Option B.