Question

Tangents are drawn to the hyperbola $4{x^2} - {y^2} = 36$ at the point P and Q. If these tangents intersect at the point$T\left( {0,3} \right)$. Then find the area (in square units) of $\Delta PTQ$.
(A) $60\sqrt 3 $
(B) $36\sqrt 5 $
(C) $45\sqrt 5 $
(D) $54\sqrt 3 $

Answer 1

Hint: Firstly use the point equation of tangent of the hyperbola to find the equation of the tangent at the point $\left( {{x_1},{y_1}} \right)$ for the given hyperbola. Now substitute the coordinates $T\left( {0,3} \right)$ into the equation. This will give you the value of y coordinate. Find two coordinates P and Q by substituting the value of y in the equation of a hyperbola. Now, find the area of the triangle made by these points P, T and Q.

Complete step-by-step answer:
It is given that the hyperbola $4{x^2} - {y^2} = 36 \Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{36}} = 1$ has tangents that pass through a common point $T\left( {0,3} \right)$.
Let us write the equation for the tangent of the hyperbola that passes through point$\left( {{x_1},{y_1}} \right)$:
$ \Rightarrow \dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$
So, we can use to write the equation for hyperbola $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{36}} = 1$ as:
$ \Rightarrow \dfrac{{x{x_1}}}{9} - \dfrac{{y{y_1}}}{{36}} = 1$
Now, for the tangent to pass through a point $T\left( {0,3} \right)$
$ \Rightarrow \dfrac{{{x_1} \times 0}}{9} - \dfrac{{{y_1} \times 3}}{{36}} = 1 \Rightarrow {y_1} = - 12$
So, we can find ${x_1}$ by putting the value of ${y_1}$ in the equation of the hyperbola
$ \Rightarrow \dfrac{{{x_1}^2}}{9} - \dfrac{{{y_1}^2}}{{36}} = 1 \Rightarrow \dfrac{{{x_1}^2}}{9} - \dfrac{{{{\left( { - 12} \right)}^2}}}{{36}} = 1$
This can be further solved by taking square root on both sides:
$ \Rightarrow \dfrac{{{x_1}^2}}{9} - \dfrac{{{{\left( { - 12} \right)}^2}}}{{36}} = 1 \Rightarrow {x_1}^2 = 9\left( {1 + \dfrac{{144}}{{36}}} \right) = 45$
$ \Rightarrow {x_1} = \pm 3\sqrt 5 $
Thus, we get the coordinates of the point $\left( {{x_1},{y_1}} \right)$ as$\left( { \pm 3\sqrt 5 , - 12} \right)$. That means there are two possibilities for this point.
Therefore, the coordinates to the point P and Q are $\left( {3\sqrt 5 , - 12} \right)$ and $\left( { - 3\sqrt 5 , - 12} \right)$
So, in the triangle$\Delta PTQ$, we have$T\left( {0,3} \right)$,$P\left( {3\sqrt 5 , - 12} \right)$ and$Q\left( { - 3\sqrt 5 , - 12} \right)$. And we have to find the area of this triangle. The point T lies on the y-axis at a distance of $3$ from origin and y coordinates for P and Q are $12$ units down to the origin. Also, the x coordinates are at the same distance from the y-axis. Therefore, we can conclude that: $\Delta PTQ$has an altitude of $15units$ and base (PQ) of $3\sqrt 5 + 3\sqrt 5 = 6\sqrt 5 units$
$ \Rightarrow $ Area of triangle $\Delta PTQ = \dfrac{1}{2} \times 15 \times 6\sqrt 5 = 45\sqrt 5 sq.units$

Hence, the option (C) is the correct answer.

Note: An alternative approach to the problem can be the use of determinants while calculating the area of the triangle. The area of the triangle $\Delta PTQ$ can be written as $\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
  {3\sqrt 5 }&{ - 12}&1 \\
  { - 3\sqrt 5 }&{ - 12}&1 \\
  0&3&1
\end{array}} \right|$