If the tangent at a point P on the parabola $y^2=3x$ is parallel to the line $x+2y=1$ and the tangents at the points Q and R on the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ are perpendicular to the line $x-y=2$, then the area of the triangle PQR is :

<p>$P \equiv \left( {{A \over {{m^2}}},{{2A} \over m}} \right)$ where $\left( {A = {3 \over 4},m = {{ - 1} \over 2}} \right)$</p> <p>& $$Q,R = \left( { \mp \,{{{a^2}{m_1}} \over {{a^2}m_1^2 + {b^2}}},{{ \mp \,.\,{b^2}} \over {\sqrt {{a^2}m_1^2 + {b^2}} }}} \right)$$</p> <p>Where ${a^2} = 4,{b^2} = 1$ and ${m_1} = 1$</p> <p>$\therefore$ $P \equiv (3, - 3)$</p> <p>$Q \equiv \left( {{{ - 4} \over {\sqrt 5 }},{{ - 1} \over {\sqrt 5 }}} \right)$ & $R\left( {{4 \over {\sqrt 5 }},{1 \over {\sqrt 5 }}} \right)$</p> <p>Area $$ = {1 \over 2}\left| {\matrix{ 3 & { - 3} & 1 \cr {{{ - 4} \over {\sqrt 5 }}} & {{{ - 1} \over {\sqrt 5 }}} & 1 \cr {{4 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} & 1 \cr } } \right| = {1 \over {10}}\left| {\matrix{ 3 & { - 3} & 1 \cr { - 4} & { - 1} & {\sqrt 5 } \cr 0 & 0 & {2\sqrt 5 } \cr } } \right|$$</p> <p>$= {{2\sqrt 5 } \over {10}}( - 15) = 3\sqrt 5$</p>