Question

TP and TQ are tangents to the parabola and the normals at P and Q meet at a point R on the curve ; prove that the centre of the circle circumscribing the triangle TPQ lies on the parabola $2{{y}^{2}}=a\left( x-a \right)$\[\]

Answer 1

Hint: We know that the parametric form of any point of the general parabola ${{y}^{2}}=4ax$ is $\left( a{{t}^{2}},2at \right)$. We assume $P\left( {{x}_{1}},{{y}_{1}} \right)$ and $Q\left( {{x}_{2}},{{y}_{2}} \right)$ ad find $T\left( {{x}_{3}},{{y}_{3}} \right)$ in a parametric form and find the equation of the chord using two-point form $ax+by+c=0$. We use the equation of circle $\left( x-{{x}_{1}} \right)\left( x-{{x}_{1}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)+m\left( ax+by+c \right)=0$ and compare the center of the circle to the center $\left( h,k \right)$. We establish a relation between $h,k$ to find the locus. \[\]

Complete step by step answer:

We are given in the question that TP and TQ are tangents to the parabola and the normals at P and Q meet at a point R on the curve. We are asked to prove that the center of the circle circumscribing the triangle TPQ lies on the parabola$2{{y}^{2}}=a\left( x-a \right)$.
We know the parametric form of any point of the general parabola ${{y}^{2}}=4ax$ is $\left( a{{t}^{2}},2at \right)$ ad where $t$ is the varying parameter .
 We assign the co-ordinates of the point of contact P and Q as $P\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right)$ and $Q\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)$ . We know that the point of intersection of tangents drawn at $P\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right)$ and $Q\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)$ is $\left( a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right)$which is the co-ordinates of T. Let us denote the centre of the circle as $\left( h,k \right)$ and we have find its locus. \[\]
We use two-point form and find the equation of the chord PQ as
\[\begin{align}
  & y-2a{{t}_{2}}=\dfrac{2a\left( {{t}_{2}}-{{t}_{1}} \right)}{{{t}_{2}}^{2}-{{t}_{1}}^{2}} \\
 & \Rightarrow \left( {{t}_{1}}+{{t}_{2}} \right)\left( y-2a{{t}_{1}} \right)=2a\left( x-a{{t}_{1}}^{2} \right) \\
 & \Rightarrow 2x-\left( {{t}_{1}}+{{t}_{2}} \right)y+2a{{t}_{1}}{{t}_{2}}=0 \\
\end{align}\]
We know that the equation of circle though parametric points $P\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right)$ ,$Q\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)$ and the chord $2x-\left( {{t}_{1}}+{{t}_{2}} \right)y+2a{{t}_{1}}{{t}_{2}}=0$ is given by
\[\left( x-a{{t}_{1}}^{2} \right)\left( x-a{{t}_{2}}^{2} \right)+\left( y-2a{{t}_{1}} \right)\left( y-2a{{t}_{2}} \right)+m\left( 2x-\left( {{t}_{1}}+{{t}_{2}} \right)y+2a{{t}_{1}}{{t}_{2}} \right)=0...(1)\]
where $m$ is ay constant. We know that $T\left( a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right)$ lies in the above circle and so we put T in above equation and get
$m=-a\left( {{t}_{1}}{{t}_{2}}+1 \right)$
We simplify equation(1) and get ,
\[{{x}^{2}}+{{y}^{2}}-x\left( a{{t}_{1}}^{2}+a{{t}_{2}}^{2}-2m \right)-y\left( 2a{{t}_{1}}+2a{{t}_{2}}+\left( {{t}_{1}}+{{t}_{2}} \right)m \right)+{{a}^{2}}{{t}_{1}}^{2}{{t}_{2}}^{2}+4{{a}^{2}}{{t}_{1}}{{t}_{2}}+2a{{t}_{1}}{{t}_{2}}k=0\]
We compare the above equation with general equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ and compare the centre as $\left( -g,-f \right)$ with denoted centre $\left( h,k \right)$. We use the value of $m=-a\left( {{t}_{1}}{{t}_{2}}+1 \right)$ and compare the abscissas to have
\[\begin{align}
  & 2h=a{{t}_{1}}^{2}+a{{t}_{2}}^{2}-2m \\
 & \Rightarrow 2h=a{{t}_{1}}^{2}+a{{t}_{2}}^{2}-2\left( -a\left( {{t}_{1}}{{t}_{2}}+1 \right) \right) \\
 & \Rightarrow 2h=a{{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}+2a...(2) \\
\end{align}\]
Similarly we compare the ordinates to have,
\[\begin{align}
  & 2k=2a{{t}_{1}}+2a{{t}_{2}}+\left( {{t}_{1}}+{{t}_{2}} \right)m \\
 & \Rightarrow 2k=2a{{t}_{1}}+2a{{t}_{2}}+\left( {{t}_{1}}+{{t}_{2}} \right)\left( -a\left( {{t}_{1}}{{t}_{2}}+1 \right) \right) \\
 & \Rightarrow 2k=a\left( 2\left( {{t}_{1}}+{{t}_{2}} \right)-3\left( {{t}_{1}}+{{t}_{2}} \right) \right) \\
 & \Rightarrow 2k=-a\left( {{t}_{1}}+{{t}_{2}} \right) \\
 & \Rightarrow {{t}_{1}}+{{t}_{2}}=\dfrac{-2k}{a} \\
\end{align}\]
We put the above value in equation (2) to get,
\[\begin{align}
  & 2h=a{{\left( \dfrac{-2k}{a} \right)}^{2}}+2a \\
 & \Rightarrow 2ah=4{{k}^{2}}+2{{a}^{2}} \\
 & \Rightarrow 2{{k}^{2}}=a\left( h-a \right) \\
\end{align}\]
So the locus of $\left( h,k \right)$ is
$2{{y}^{2}}=a\left( x-a \right)$\[\]
Note:
The general equation of any parabola is ${{y}^{2}}=4ax$ and the equation of tangent at any point on the parabola is given by $y=mx+\dfrac{a}{m}$. The equation of tangent at any point $\left( {{x}_{1}},{{y}_{1}} \right)$ is $y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)$