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Question: Let \(PQ\) be a double ordinate of the parabola, \({y^2} = - 4x\), where \(P\) lies in the second qu...

Let PQPQ be a double ordinate of the parabola, y2=−4x{y^2} = - 4x, where PP lies in the second quadrant. If RR divides PQPQ in the ratio 2 : 1, then the locus of RR is
A. 3y2=−2x B. 9y2=4x C. 9y2=−4x D. 3y2=2x  {\text{A}}{\text{. 3}}{y^2} = - 2x \\\ {\text{B}}{\text{. }}9{y^2} = 4x \\\ {\text{C}}{\text{. }}9{y^2} = - 4x \\\ {\text{D}}{\text{. }}3{y^2} = 2x \\\

Explanation

Solution

Hint: In this question, firstly we have to analyse the question and understand the concept of parabola. We have to identify the ratio in which the ordinate is divided and then find its locus by checking all the options one by one.

Complete step-by-step answer:

Now assume the coordinates of points {\text{P,Q & R}} as
P(−at12,2at1),Q(−at12,2at1){\text{P}}\left( { - at_1^2,2a{t_1}} \right),Q\left( { - at_1^2,2a{t_1}} \right) and R(h,k){\text{R}}\left( {h,k} \right)
As R{\text{R}} divides PQ{\text{PQ}}in the ratio 2:1 according to question, we take the coordinates of P & Q in that ratio and equate it with the coordinates of R{\text{R}}
Then, step 1 is
h=−1at12−2at122+1=−at12h = \dfrac{{ - 1at_1^2 - 2at_1^2}}{{2 + 1}} = - at_1^2
and step 2 is
k=1×2at1−2×2at12+1=−2at13k = \dfrac{{1 \times 2a{t_1} - 2 \times 2a{t_1}}}{{2 + 1}} = \dfrac{{ - 2a{t_1}}}{3}
Now from step 2 we gather the value of t1{t_1} and put it in step 1. So, we get
h=−a(3k−2a)2=−9k24h = - a{\left( {\dfrac{{3k}}{{ - 2a}}} \right)^2} = - \dfrac{{9{k^2}}}{4}
⇒9k2=−4h\Rightarrow 9{k^2} = - 4h or we can put x&yx\& y in place of h&kh\& k
9y2=−4x9{y^2} = - 4x
Hence, option C is the correct answer.

Note: Whenever we face such types of questions the key point to remember is that any chord that is perpendicular to the axis of parabola is called a double ordinate. So, we have to assume the points on parabola and equate them to the point on double ordinate and then check our options and find the right answer.