Conic-Sections Question 565
Question: A line PQ meets the parabola $ y^{2}-4ax $ in R such that PQ is bisected at R. if the coordinates of P are $ (x_1,y_1) $ then the locus of Q is the parabola
Options:
A) $ {{(y+y_1)}^{2}}=8a(x+x_1) $
B) $ {{(y-y_1)}^{2}}=8a(x+x_1) $
C) $ {{(y+y_1)}^{2}}=8a(x-x_1) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let the coordinates of Q be $ (h,k) $ . Since the point R lies on the parabola. Let it coordinates be $ (at^{2},2at). $ Since R is midpoint of PQ,
$ \therefore at^{2}=\frac{x_1+h}{2} $ and $ 2at=\frac{y_1+k}{2} $
$ \Rightarrow t^{2}=\frac{x_1+h}{2a} $ And $ t=\frac{y_1+k}{4a} $ Equating the two values of t, we get $ {{( \frac{y_1+k}{4a} )}^{2}}=\frac{x_1+h}{2a}\Rightarrow {{(y_1+k)}^{2}}=8a(x_1+h) $ Hence, locus of Q (h, k) is $ {{(y+y_1)}^{2}}=8a(x+x_1) $
BETA
