Question

The normals at three points P, Q, R of the parabola ${{y}^{2}}=4ax$ meet in $\left( h,k \right)$. The centroid of triangle PQR lies on $$$$
A.x=0$$$$$ B.y=0 C.$x=-a
D.$y=-a$$$$$

Answer 1

Hint: We use the general equation of normal of given rightward parabola $y=mx-2am-a{{m}^{3}}$ and use the satisfaction of the point $\left( h,k \right)$ and also the coordinates feet of the normals given as P,Q,R in the form $\left( a{{m}^{2}},-2am \right)$.We denote the roots as ${{m}_{1}},{{m}_{2}},{{m}_{3}}$ which are slopes of the normals . We then use sum of roots of and product of roots of a cubic polynomial to find ${{m}_{1}}+{{m}_{2}}+{{m}_{3}},{{m}_{1}}{{m}_{2}}+{{m}_{2}}{{m}_{3}}+{{m}_{3}}{{m}_{1}}$. We use it to find the centroid of the triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.$$$$

Complete step-by-step answer:

We know that the standard equation of a normal at any point with slope $m$ of the rightward parabola ${{y}^{2}}=4ax$ is given by

$y=mx-2am-a{{m}^{3}}$

The foot of the perpendicular of the above normal is given by $\left( a{{m}^{2}},-2am \right)$. Since we are given there are three normals; let us denote their slopes as ${{m}_{1}},{{m}_{2}},{{m}_{3}}$. We are also given the normals meet at $\left( h,k \right)$.So by satisfaction of $\left( h,k \right)$ equation of normal is

$\begin{aligned} & \Rightarrow k=mh-2am-a{{m}^{3}} \\\ & \Rightarrow a{{m}^{3}}+\left( 2a-h \right)m+k=0 \\\ \end{aligned}$

The roots of the cubic equation will be the slopes of normals that is ${{m}_{1}},{{m}_{2}},{{m}_{3}}$. We are given that the normals are drawn at P, Q and R. So using the coordinates of feet of the perpendicular of normal $\left( a{{m}^{2}}.-2am \right)$ we can assign

$P\left( am_{_{1}}^{2},-2a{{m}_{1}} \right),Q\left( am_{_{2}}^{2},-2a{{m}_{2}} \right),R\left( am_{_{3}}^{2},-2a{{m}_{3}} \right)$

So by sum of roots of cubic polynomial we have;

$\begin{aligned} & {{m}_{1}}+{{m}_{2}}+{{m}_{3}}=\dfrac{0}{a}=0 \\\ & {{m}_{1}}{{m}_{2}}+{{m}_{2}}{{m}_{3}}+{{m}_{3}}{{m}_{1}}=\dfrac{2a-h}{a} \\\ \end{aligned}$

Let us denote the centroid of the triangle PQR as $G$. The $x-$coordinate of the centre is given by

$\dfrac{am_{1}^{2}+am_{2}^{2}+am_{3}^{2}}{3} \\\ \Rightarrow \dfrac{a\left( m_{1}^{2}+m_{2}^{2}+m_{3}^{2} \right)}{3} \\\ \Rightarrow a\dfrac{{{\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)}^{2}}-2\left( {{m}_{1}}{{m}_{2}}+{{m}_{2}}{{m}_{3}}+{{m}_{3}}{{m}_{1}} \right)}{3} \\\ \Rightarrow a\dfrac{{{0}^{2}}-2\dfrac{\left( 2a-h \right)}{a}}{3}=\dfrac{2}{3}\left( h-2a \right) \\\$

Let us find the $y-$coordinate of $G$. We have;

$\begin{aligned} & \dfrac{-2a{{m}_{1}}+\left( -2a{{m}_{2}} \right)+\left( -2a{{m}_{3}} \right)}{3} \\\ & \Rightarrow \dfrac{-2a\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)}{3} \\\ & \Rightarrow \dfrac{-2a\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)}{3}=0 \\\ \end{aligned}$

So the coordinates of $G$ are $G\left( \dfrac{2}{3}\left( h-2a \right),0 \right)$. So the $y-$coordinate of G will always be zero which means G will always lie on the $x-$axis whose equation is $y=0$.

So, the correct answer is “Option A”.

Note: We note that the general form of the cubic equation is given by $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$. If the roots of the cubic equation are ${{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}}$ then the sum of the roots is given by ${{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}=\dfrac{-b}{a}$, the sum of the roots taking product of two at a time is given by ${{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{2}}{{\alpha }_{3}}+{{\alpha }_{3}}{{\alpha }_{1}}=\dfrac{c}{a}$ and the product of the roots is given by ${{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}=\dfrac{-d}{a}$. We can directly solve if we know that the centroid of the triangle made by feet of three normals always lies on the axis of the parabola and the axis of given parabola ${{y}^{2}}=4ax$is $y=0$.