Question

The tangents at three points $A,B,C$ on the parabola $y^2=4x$, taken in pairs intersect at the points $P,Q$ and $R.$ If $△$, ${△}^{\prime }$ be the areas of the triangles $ABC$ and $PQR$ respectively, then

• A. $△={△}^{\prime }$
• B. $2{△}^{\prime }=△$
• C. $2△={△}^{\prime }$
• D. none of these

Answer 1

The correct option is B $2{△}^{\prime }=△$

Let the coordinates of $A,B,C$ be $({t_i}^2,2t_i),i=1,2,3$ respectively.

The tangents at $A$ and $B$ are $t_{1}y=x+{t_1}^2$ and $t_{2}y=x+{t_2}^2$

which intersect at $x=t_1{t}_2,y=t_1+t_2$

So the vertices are $P(t_1{t}_2,t_1+t_2),Q(t_2{t}_3,t_2+t_3)$ and $R(t_1{t}_3,t_1+t_3)$

$\Delta =\left|(t_1-t_2)(t_2-t_3)(t_3-t_1)\right|$

${\Delta }^{\prime }=\left|\frac{1}{2}\left|\begin{array}{ccc}{t}_1{t}_2& t_1{+t}_2& 1\\ t_2{t}_3& t_2{+t}_3& 1\\ t_3{t}_1& t_3{+t}_1& 1\end{array}\right|\right|=\left|\frac{1}{2}\left|\begin{array}{ccc}(t_1-t_3)t_2& t_1-t_3& 0\\ (t_2-t_1)t_3& t_2-t_1& 0\\ t_3{t}_1& t_3+t_1& 1\end{array}\right|\right|$

$=\left|\frac{1}{2}(t_1-t_3)(t_2-t_1)(t_2-t_3)\right|\Rightarrow \Delta =2{\Delta }^{\prime }$