This problem is a classic application of the chord of contact concept in coordinate geometry, specifically for a hyperbola. We will leverage this powerful tool to efficiently find the points of tangency and subsequently the area of the triangle formed.

---

1. Fundamental Concept: The Chord of Contact

When two tangents are drawn from an external point $T(x_1, y_1)$ to a conic section, the line segment connecting the two points of tangency (let's call them P and Q) is known as the chord of contact. The remarkable utility of this concept lies in the fact that we can directly determine the equation of this line segment PQ without needing to find the equations of the individual tangents or the coordinates of P and Q first.

For a hyperbola given by the general equation $Ax^2 - By^2 = C$, the equation of the chord of contact from an external point $T(x_1, y_1)$ is given by: $Axx_1 - Byy_1 = C$ This formula is a direct application of the "T=0" rule, where $T$ represents the equation of the tangent if the point $(x_1, y_1)$ were on the conic.

---

2. Problem Setup: Identifying the Given Information

Let's clearly lay out the information provided in the problem statement:

• The equation of the hyperbola is: $4x^2 - y^2 = 36$.
• The external point from which the tangents are drawn is $T(x_1, y_1) = T(0, 3)$.
• P and Q are the points on the hyperbola where the tangents touch it.
• Our ultimate goal is to calculate the area of $\triangle PTQ$.

---

3. Step 1: Determine the Equation of the Chord of Contact (Line PQ)

Our first strategic move is to find the equation of the line segment PQ, which is the chord of contact. This line will serve as the base of our triangle $\triangle PTQ$.

We apply the chord of contact formula $Axx_1 - Byy_1 = C$. From the hyperbola equation $4x^2 - y^2 = 36$, we identify:

• $A = 4$ (coefficient of $x^2$)
• $B = 1$ (coefficient of $y^2$; note that $y^2 = 1 \cdot y^2$)
• $C = 36$ (the constant term) The external point is $(x_1, y_1) = (0, 3)$.

Substitute these values into the formula: $4x(0) - 1y(3) = 36$ $0 - 3y = 36$ $-3y = 36$ Now, divide both sides by $-3$ to solve for $y$: $y = \frac{36}{-3}$ $y = -12$ This is the equation of the chord of contact, which is the line PQ. Explanation: By using the chord of contact formula, we efficiently obtain the equation of the line on which P and Q lie. This bypasses the more laborious process of finding individual tangent equations and their intersection points with the hyperbola. The result $y=-12$ indicates that PQ is a horizontal line.

---

4. Step 2: Find the Coordinates of the Points P and Q

The points P and Q are the points of tangency, which means they are the intersection points of the chord of contact ($y = -12$) and the hyperbola ($4x^2 - y^2 = 36$). To find their coordinates, we substitute the equation of the chord of contact into the hyperbola's equation.

Substitute $y = -12$ into $4x^2 - y^2 = 36$: $4x^2 - (-12)^2 = 36$ $4x^2 - 144 = 36$ Next, isolate the $x^2$ term: $4x^2 = 36 + 144$ $4x^2 = 180$ Divide by 4: $x^2 = \frac{180}{4}$ $x^2 = 45$ Take the square root of both sides to find $x$: $x = \pm\sqrt{45}$ To simplify the radical, we factor 45 into $9 \times 5$: $x = \pm\sqrt{9 \times 5}$ $x = \pm 3\sqrt{5}$ So, the two x-coordinates are $3\sqrt{5}$ and $-3\sqrt{5}$. Since we substituted $y=-12$, the y-coordinate for both points P and Q is $-12$. Therefore, the coordinates of the points of tangency are: $P = (-3\sqrt{5}, -12)$ $Q = (3\sqrt{5}, -12)$ (Note: The specific assignment of P and Q to the positive or negative x-value is arbitrary and does not affect the final area calculation.) Explanation: By finding the intersection points of the chord of contact with the hyperbola, we precisely locate the vertices P and Q of our triangle. The fact that both points share the same y-coordinate ($y=-12$) confirms that PQ is indeed a horizontal line, as determined in Step 1.

---

5. Step 3: Calculate the Length of the Base of the Triangle (PQ)

The triangle we are interested in is $\triangle PTQ$. We now have the coordinates of all three vertices: $T(0, 3)$, $P(-3\sqrt{5}, -12)$, and $Q(3\sqrt{5}, -12)$. We can consider the line segment PQ as the base of the triangle. Since P and Q have the same y-coordinate, the length of PQ is simply the absolute difference of their x-coordinates.

Length of PQ (Base) $= |x_Q - x_P|$ $= |3\sqrt{5} - (-3\sqrt{5})|$ $= |3\sqrt{5} + 3\sqrt{5}|$ $= |6\sqrt{5}|$ $= 6\sqrt{5} \text{ units}$ Explanation: This is a direct application of the distance formula for two points that lie on a horizontal line. This length is a crucial component for calculating the area of $\triangle PTQ$.

---

6. Step 4: Calculate the Height of the Triangle (from T to PQ)

The height of $\triangle PTQ$ is the perpendicular distance from the vertex T to its opposite side, the base PQ.

• The coordinates of vertex T are $(0, 3)$.
• The equation of the line PQ is $y = -12$.

Since PQ is a horizontal line ($y = -12$), the perpendicular distance from T to PQ is simply the absolute difference between the y-coordinate of T and the y-coordinate of the line PQ. Let M be the foot of the perpendicular from T to PQ. The coordinates of M would be $(0, -12)$, as T has an x-coordinate of 0, which lies between the x-coordinates of P and Q.

Height (TM) $= |y_T - y_{PQ}|$ $= |3 - (-12)|$ $= |3 + 12|$ $= |15|$ $= 15 \text{ units}$ Explanation: The height of a triangle is always measured perpendicularly from a vertex to its base. Because the base PQ is a horizontal line, the height is the vertical distance between the y-coordinate of T and the y-coordinate of the line PQ. This simplifies the calculation significantly compared to using the general distance formula from a point to a line.

---

7. Step 5: Calculate the Area of $\triangle PTQ$

Now that we have the length of the base (PQ) and the height (TM), we can calculate the area of the triangle using the standard formula: Area of a triangle $= \frac{1}{2} \times \text{base} \times \text{height}$

Substitute the values we found: $\text{Area}(\triangle PTQ) = \frac{1}{2} \times (\text{Length of PQ}) \times (\text{Height TM})$ $\text{Area}(\triangle PTQ) = \frac{1}{2} \times (6\sqrt{5}) \times (15)$ $= (3\sqrt{5}) \times 15$ $= 45\sqrt{5} \text{ square units}$

---

Important Tips and Common Pitfalls for JEE Aspirants:

• Master the Chord of Contact Formula: This formula ($T=0$ or $S_1=0$) is a cornerstone for problems involving tangents from an external point to all conic sections (parabola, ellipse, hyperbola, circle). Knowing it thoroughly saves immense time.
• Algebraic Precision is Key: Be extremely careful with signs, especially when squaring negative numbers (e.g., $(-12)^2 = 144$) and when calculating differences for distances (e.g., $3 - (-12) = 15$). A single sign error can lead to a completely wrong answer.
• Simplifying Radicals: Always simplify square roots to their simplest form (e.g., $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$). This makes calculations cleaner and ensures your answer matches the options provided.
• Geometric Visualization: A quick mental sketch or rough drawing of the situation can provide valuable insights. Recognizing that PQ is a horizontal line and that T is vertically aligned with the midpoint of PQ (since T is on the y-axis and the hyperbola is symmetric about the y-axis) significantly simplifies the height calculation.
• Avoid Lengthy Alternatives: While it's theoretically possible to find the equations of the two tangents, then their intersection points with the hyperbola, and then calculate the area using the determinant method (or Shoelace formula), this approach is significantly more cumbersome and time-consuming. The chord of contact method is designed for efficiency.

---

Summary and Key Takeaway:

This problem effectively tests your understanding and application of the chord of contact concept for hyperbolas. The most efficient and elegant strategy involves:

1. Using the chord of contact formula to find the equation of the line segment PQ.
2. Finding the coordinates of the points P and Q by intersecting this line with the hyperbola.
3. Calculating the length of PQ (the base of the triangle) and the perpendicular distance from T to PQ (the height of the triangle).
4. Applying the standard area formula for a triangle.

This systematic approach avoids complex calculations and provides a straightforward path to the solution.

The final answer is $\boxed{45\sqrt 5}$.