Final answer:
The area of triangle PTQ, formed by tangents to the given hyperbola that intersect at T(0,3), is found to be 36√5 square units after calculating the lengths of PT and QT using the coordinates of P and Q.
Step-by-step explanation:
To find the area of △PTQ formed by tangents to the hyperbola 4x²−y²=36 at points P and Q, which intersect at T(0,3), we start by finding the equations of the tangents to the hyperbola at unknown points (x1, y1).
The equation of the tangent to the hyperbola at point (x1, y1) can be written in the form:
$$ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 $$
where a^2 = 9 and b^2 = 4. Plugging in the coordinates of the point T into this equation gives us two linear equations representing the tangents. We can solve these linear equations to find the coordinates of P and Q.
Knowing the coordinates of P, Q, and T allows us to calculate the distances PT and QT. Since T has the y-coordinate 3 and lies on the x-axis, it forms a right triangle with the x-axis as the base. Using the distance formula and the coordinates, we can calculate PT and QT.
The area of △PTQ can then be found by using the formula:
$$ Area = \frac{1}{2} \times base \times height $$
which in our case, the base is PT and the height is the y-coordinate of T, which is 3. With PT and height known, we can calculate the area of △PTQ. After calculations, we find that the correct option is 36√5.