Question

A tangent to the parabola${{y}^{2}}=4ax$meets the axes at A and B. The locus of midpoint of AB is:
a) ${{y}^{2}}+2ax=0$
b) ${{y}^{2}}-2ax=0$
c) ${{y}^{2}}+ax=0$
d) $2{{y}^{2}}+ax=0$

Answer 1

Hint: Write the equation of the tangent on the parabola $ {{y}^{2}}=4ax $ at a parametric point $ \left( a{{t}^{2}},2at \right) $ .Then find the points on x and y axis where this tangent is intersecting. The intersecting points will be A and B then find the midpoint of A and B.

Complete step-by-step answer:
Let us take a parametric point $ \left( a{{t}^{2}},2at \right) $ which lie on the parabola $ {{y}^{2}}=4ax $ then we are going to write the equation of a tangent at this parametric point.
The slope of the tangent of the parabola $ {{y}^{2}}=4ax $ is calculated below:
  $ {{y}^{2}}=4ax $
Taking derivative with respect to x on both the sides will get:
  $ \begin{align}
  & 2y\dfrac{dy}{dx}=4a \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y} \\
\end{align} $
Now, substituting the parametric point $ \left( a{{t}^{2}},2at \right) $ in the above equation we get the slope as:
  $ \begin{align}
  & \dfrac{dy}{dx}=\left( \dfrac{2a}{2at} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{t} \\
\end{align} $
The equation of a tangent at the parametric point with the slope $ \dfrac{1}{t} $ is:
  $ y-2at=\dfrac{1}{t}\left( x-a{{t}^{2}} \right) $
When the above equation cut x axis then y = 0 and the coordinate of x is:
  $ \begin{align}
  & 0-2at=\dfrac{1}{t}\left( x-a{{t}^{2}} \right) \\
 & \Rightarrow -2a{{t}^{2}}=x-a{{t}^{2}} \\
 & \Rightarrow x=-a{{t}^{2}} \\
\end{align} $
When the above equation cut y axis then x = 0 and the coordinate of y is:
  $ \begin{align}
  & y-2at=\dfrac{1}{t}\left( 0-a{{t}^{2}} \right) \\
 & \Rightarrow y=2at-at \\
 & \Rightarrow y=at \\
\end{align} $
From the above calculations, coordinates of $ A\left( -a{{t}^{2}},0 \right) $ and $ B\left( 0,at \right) $ .
What we have described above is shown through the below diagram:

In the above figure, you can see a parabola $ {{y}^{2}}=4ax $ on which a tangent is drawn at point $ P\left( a{{t}^{2}},2at \right) $ and the tangent intersects X and Y axis at $ A\left( -a{{t}^{2}},0 \right) $ and $ B\left( 0,at \right) $ respectively.
The x and y coordinates of the midpoint of A and B is equal to:
  $ \left( -\dfrac{a{{t}^{2}}}{2},\dfrac{at}{2} \right) $
So, from the above expression $ x=-\dfrac{a{{t}^{2}}}{2} $ and $ y=\dfrac{at}{2} $ .
  $ y=\dfrac{at}{2} $
Squaring both the sides will give:
  $ \begin{align}
  & {{y}^{2}}=\dfrac{{{a}^{2}}{{t}^{2}}}{4} \\
 & \Rightarrow 4{{y}^{2}}={{a}^{2}}{{t}^{2}} \\
 & \Rightarrow {{t}^{2}}=\dfrac{4{{y}^{2}}}{{{a}^{2}}} \\
\end{align} $
Substituting the value of $ {{t}^{2}} $ in $ x=-\dfrac{a{{t}^{2}}}{2} $ we get,
  $ \begin{align}
  & x=-\dfrac{a\left( \dfrac{4{{y}^{2}}}{{{a}^{2}}} \right)}{2} \\
 & \Rightarrow -2ax=4{{y}^{2}} \\
 & \Rightarrow 2{{y}^{2}}+ax=0 \\
\end{align} $
Hence, the locus of the midpoint of A and B is $ 2{{y}^{2}}+ax=0 $ .
Hence, the correct option is (d).

Note: The locus of a point is a relation between x and y which is holding one or more conditions like we have to find the locus of a midpoint of A and B.