Question

Let P be the point (1,0) and Q a point of the locus \[{{y}^{2}}=8x\]. The locus of the midpoint of PQ is
(a) \[{{x}^{2}}+4y+2=0\]
(b) \[{{x}^{2}}-4y+2=0\]
(c) \[{{y}^{2}}-4x+2=0\]
(d) \[{{y}^{2}}+4x+2=0\]

Answer 1

Hint: To find the locus of points joining one point on the parabola to another point, assume any point on the curve and find the midpoint of two given points. Eliminate the variables to find the equation of locus.

Complete step-by-step answer:

We have the equation of parabola as \[{{y}^{2}}=8x\]. We have a point \[\left( 1,0 \right)\] outside the parabola. We have to find the locus of midpoint joining one point on the parabola to the point (1,0).
We know that any general point on the parabola \[{{y}^{2}}=4ax\] has form \[\left( a{{t}^{2}},2at \right)\]. Let us assume that these are the coordinates of point Q.

Substituting \[a=2\] in the above equation, we get \[Q\left( 2{{t}^{2}},4t \right)\] as the point on the parabola.
We know that the middle point of two points of the form \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\].
Substituting \[{{x}_{1}}=1,{{y}_{1}}=0,{{x}_{2}}=a{{t}^{2}},{{y}_{2}}=2at\] in the above equation, we have \[\left( \dfrac{1+a{{t}^{2}}}{2},\dfrac{0+2at}{2} \right)\] as the middle point of the points \[P(1,0)\] and \[Q\left( a{{t}^{2}},2at \right)\].
Thus, we have \[\left( \dfrac{1+2{{t}^{2}}}{2},2t \right)\] as the middle points of the points \[P(1,0)\] and \[Q\left( 2{{t}^{2}},4t \right)\].
To find the locus of middle points, we will assume \[x=\dfrac{1+2{{t}^{2}}}{2}.....\left( 1 \right)\] and \[y=2t.....\left( 2 \right)\].
Rewriting equation (2) in terms of t by rearranging the terms, we have \[\dfrac{y}{2}=t.....\left( 3 \right)\].
Substituting equation (3) in equation (1), we get \[x=\dfrac{1+2{{\left( \dfrac{y}{2} \right)}^{2}}}{2}\].
Rearranging the terms, we get \[2x=1+2{{\left( \dfrac{y}{2} \right)}^{2}}\].
Further solving the above equation, we get
\[\begin{align}
  & \Rightarrow 2x=1+\dfrac{{{y}^{2}}}{2} \\
 & \Rightarrow 4x=2+{{y}^{2}} \\
 & \Rightarrow {{y}^{2}}-4x+2=0 \\
\end{align}\]
Hence, the locus of middle point of point P(1,0) and point Q on the parabola is \[{{y}^{2}}-4x+2=0\], which is option (c).

Note: It’s necessary to use the midpoint formula to find the midpoint of two points. We can also solve this question by assuming any general point on the parabola instead of taking the general point in parametric form.