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

Options

A. `36sqrt5`
B. `45sqrt5`
C. `54sqrt3`
D. `60sqrt3`

Explanation

To find the area of triangle PTQ where T is the intersection point of tangents drawn to the hyperbola \(4x^2 - y^2 = 36\) at points P and Q, we can follow these steps:

### Step 1: Rewrite the Hyperbola in Standard Form
The given hyperbola is:
\[ 4x^2 - y^2 = 36 \]
Dividing the entire equation by 36, we get:
\[ \frac{x^2}{9} - \frac{y^2}{36} = 1 \]
This is the standard form of the hyperbola.

### Step 2: Determine the Equation of the Tangent
The equation of the tangent to the hyperbola at a point \((x_1, y_1)\) is given by:
\[ \frac{xx_1}{9} - \frac{yy_1}{36} = 1 \]
Since the tangents intersect at point T(0, 3), we substitute \(x = 0\) and \(y = 3\) into the tangent equation:
\[ \frac{0 \cdot x_1}{9} - \frac{3y_1}{36} = 1 \]
This simplifies to:
\[ -\frac{3y_1}{36} = 1 \implies y_1 = -12 \]
Thus, the y-coordinate of both points P and Q is \(-12\).

### Step 3: Find the x-coordinates of Points P and Q
Since the points P and Q lie on the hyperbola, we substitute \(y = -12\) into the hyperbola's equation:
\[ \frac{x^2}{9} - \frac{(-12)^2}{36} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} - \frac{144}{36} = 1 \implies \frac{x^2}{9} - 4 = 1 \]
Rearranging gives:
\[ \frac{x^2}{9} = 5 \implies x^2 = 45 \implies x = \pm 3\sqrt{5} \]
Thus, the coordinates of points P and Q are:
- \(P(-3\sqrt{5}, -12)\)
- \(Q(3\sqrt{5}, -12)\)

### Step 4: Calculate the Area of Triangle PTQ
The area of triangle PTQ can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
Substituting the coordinates \(P(-3\sqrt{5}, -12)\), \(Q(3\sqrt{5}, -12)\), and \(T(0, 3)\):
\[ \text{Area} = \frac{1}{2} \left| (-3\sqrt{5})(-12 - 3) + (3\sqrt{5})(3 - (-12)) + (0)(-12 - (-12)) \right| \]
This simplifies to:
\[ = \frac{1}{2} \left| (-3\sqrt{5})(-15) + (3\sqrt{5})(15) \right| \]
\[ = \frac{1}{2} \left| 45\sqrt{5} + 45\sqrt{5} \right| = \frac{1}{2} \left| 90\sqrt{5} \right| = 45\sqrt{5} \]

### Final Answer
The area of triangle PTQ is:
\[ \text{Area} = 45\sqrt{5} \text{ square units} \]