Question

Find the equation of tangent to parabola ${{y}^{2}}=3x$ at point P(t) and hence find the locus of the centroid of the triangle TPS, where T is the point of intersection of tangent and axis of parabola and S is the focus of parabola.

Answer 1

Hint: First we are going to draw the diagram and then we will take a parametric point P on parabola, If the equation of parabola is ${{y}^{2}}=4ax$, then the parametric point on parabola is $\left( a{{t}^{2}},2at \right)$ and then we will use the formula for tangent and then we will find T and S, after that we use the formula for finding the centroid of the triangle.

Complete step-by-step answer:

If the equation of parabola is ${{y}^{2}}=4ax$,
Then the parametric point on parabola is $\left( a{{t}^{2}},2at \right)$
By comparing ${{y}^{2}}=4ax$ with ${{y}^{2}}=3x$ we get,
The value of a = $\dfrac{3}{4}$ ,
If the equation of parabola is ${{y}^{2}}=4ax$,
Then the focus is (a,0), using this we get,
The point S$\left( \dfrac{3}{4},0 \right)$ as the focus, T is the point where tangents cuts the axis, and P$\left( \dfrac{3{{t}^{2}}}{4},\dfrac{2\times 3t}{4} \right)$ is the parametric point on parabola.
The equation of tangent at a point (h, k) on this parabola ${{y}^{2}}=4ax$ is: \[\text{ky=}\dfrac{4a(x+h)}{2}\]
Therefore the equation of tangent at point P is:
$\begin{align}
  & \dfrac{3ty}{2}=\dfrac{3\left( x+\dfrac{3{{t}^{2}}}{4} \right)}{2} \\
 & ty=x+\dfrac{3{{t}^{2}}}{4} \\
\end{align}$
Now to find point T we have to put y = 0, as it cuts x axis.
$x=-\dfrac{3{{t}^{2}}}{4}$
Therefore, the point T will be: $\left( -\dfrac{3{{t}^{2}}}{4},0 \right)$
Now we have all the three points of the triangle.
The formula for finding centroid if the three points are A(x, y) B(s,t) C(q,r) is:
$\left( \dfrac{x+s+q}{3},\dfrac{y+t+r}{3} \right)$
Now using this formula on triangle TPS we get,
$\begin{align}
  & \left( \dfrac{\dfrac{-3{{t}^{2}}}{4}+\dfrac{3{{t}^{2}}}{4}+\dfrac{3}{4}}{3},\dfrac{0+\dfrac{3t}{2}+0}{3} \right) \\
 & \left( \dfrac{1}{4},\dfrac{t}{2} \right) \\
\end{align}$
Hence this is the value of the centroid of triangle TPS.

Note: One should know the parametric form of parabola, if the equation of parabola is ${{y}^{2}}=4ax$then the parametric point on parabola is $\left( a{{t}^{2}},2at \right)$ , and then how to find the equation of tangent, and then the formula for centroid of triangle. All these formulas should be remembered or we can make mistakes. Keep in mind that the formula for tangent that we have used is only applicable when the given point is on the parabola, so it should not be used if the point is not on the parabola.