Question

Let O be the vertex and Q be any point on the parabola \[{{x}^{2}}=8y\]. If the point P divides the line segment OQ internally in the ratio \[1:3\], then the locus of P is
\[\begin{align}
  & A){{x}^{2}}=y \\
 & B){{y}^{2}}=x \\
 & C){{y}^{2}}=2x \\
 & D){{x}^{2}}=2y \\
\end{align}\]

Answer 1

Hint: We know that the vertex of the parabola \[{{x}^{2}}=4ay\]is \[O\left( 0,0 \right)\]. We know that the general point on the parabola \[{{x}^{2}}=4ay\] is \[Q\left( 2at,a{{t}^{2}} \right)\]. We know that if \[P\left( {{x}_{1}},{{y}_{1}} \right)\] and \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] are two points, then \[R\left( {{x}_{3}},{{y}_{3}} \right)\] is divided by \[P\left( {{x}_{1}},{{y}_{1}} \right)\] and \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] in the ratio \[m:n\]internally, then \[R\left( {{x}_{3}},{{y}_{3}} \right)\] is equal to \[R\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)\]. By using this concept, we can find the equation of parabola.

Complete step-by-step answer:
We know that the vertex of the parabola \[{{x}^{2}}=4ay\]is \[O\left( 0,0 \right)\].From the question, we were given that O is the vertex of the parabola.
Now we should compare \[{{x}^{2}}=8y\] with \[{{x}^{2}}=4ay\].
\[\begin{align}
  & \Rightarrow 4a=8 \\
 & \Rightarrow a=2....(1) \\
\end{align}\]
So, from equation (1) we can write the vertex of the parabola \[{{x}^{2}}=8y\] is \[O\left( 0,0 \right)\].
We know that the general point on the parabola \[{{x}^{2}}=4ay\] is \[Q\left( 2at,a{{t}^{2}} \right)\].
So, from equation (1), we can write that the general point on the parabola \[{{x}^{2}}=8y\] is \[Q\left( 4t,2{{t}^{2}} \right)\].
We know that if \[P\left( {{x}_{1}},{{y}_{1}} \right)\] and \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] are two points, then \[R\left( {{x}_{3}},{{y}_{3}} \right)\] is divided by \[P\left( {{x}_{1}},{{y}_{1}} \right)\] and \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] in the ratio \[m:n\]internally, then \[R\left( {{x}_{3}},{{y}_{3}} \right)\] is equal to \[R\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)\].

From the question, we were given that the point \[P(x,y)\] divides the line segment OQ internally in the ratio \[1:3\].
\[\begin{align}
  & \Rightarrow P(x,y)=P\left( \dfrac{\left( 1 \right)(4t)+(3)(0)}{1+3},\dfrac{(1)(2{{t}^{2}})+(3)(0)}{1+3} \right) \\
 & \Rightarrow P(x,y)=P\left( \dfrac{4t}{4},\dfrac{2{{t}^{2}}}{4} \right) \\
 & \Rightarrow P(x,y)=P\left( t,\dfrac{{{t}^{2}}}{2} \right) \\
\end{align}\]
We know that the x-coordinate of \[P(x,y)\] is equal to x and the y-coordinate of \[P(x,y)\] is equal to y.
\[\begin{align}
  & x=t....(2) \\
 & y=\dfrac{{{t}^{2}}}{2}.....(3) \\
\end{align}\]
Now let us substitute equation (2) in equation (3), then
\[\begin{align}
  & y=\dfrac{{{x}^{2}}}{2} \\
 & \Rightarrow {{x}^{2}}=2y......(4) \\
\end{align}\]
So, the correct answer is “Option D”.

Note: Students may have a misconception that if \[P\left( {{x}_{1}},{{y}_{1}} \right)\] and \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] are two points, then \[R\left( {{x}_{3}},{{y}_{3}} \right)\] is divided by \[P\left( {{x}_{1}},{{y}_{1}} \right)\] and \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] in the ratio \[m:n\] internally, then \[R\left( {{x}_{3}},{{y}_{3}} \right)\] is equal to \[R\left( \dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n},\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n} \right)\]. But we know that if \[P\left( {{x}_{1}},{{y}_{1}} \right)\] and \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] are two points, then \[R\left( {{x}_{3}},{{y}_{3}} \right)\] is divided by \[P\left( {{x}_{1}},{{y}_{1}} \right)\] and \[Q\left( {{x}_{2}},{{y}_{2}} \right)\] in the ratio \[m:n\], then \[R\left( {{x}_{3}},{{y}_{3}} \right)\] is equal to \[R\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)\]. So, this misconception should be avoided.