Let P be the point (1,0) and Q be a point on the locus `y^(2)=8x`. The locus of the midpoint of PQ is

Options

A. `P_(1)` lies on `C_(2)andQ_(1)` lies on `C_(1)`
B. `PQgemin{PP_(1),Q Q_(1)}`
C. point `P_(0)` on `C_(1)` such that `P_(0)Q_(0)lePQ` for all pairs of points (P,Q) is `((1)/(3),(10)/(9))`
D. point `Q_(0)` on `C_(2)` such that `P_(0)Q_(0)lePQ` for all pairs of points (P,Q) is `((10)/(9),(1)/(3))`

Answer

A, B

Explanation

1,2 Given that `C_(1):x^(2)=y-1`
`C_(2):y^(2)=y-1`
Here, `C_(1)andC_(2)` are symmetrical about the line y=x.
Let us take points `P(x_(1),x_(1)^(2)+1)" on "C_(1)andQ(y_(2)^(2)+1,y_(2))" on "C_(2)`.
Then image of P in y=x is `P_(1)(x_(1)^(2)+1,x_(1))" on "C_(2) "and image of Q in y=x" is "Q_(1)(y_(2),y_(2)^(2)+1)" on "C_(1)`.

Now, `PP_(1)andQQ_(1)` both are perpendicular to mirror line y=x.
Also, M is midpoint of `PP_(1). " "(becauseP_(1)` is mirror of P in y=x)
`:.PM=(1)/(2)PP_(1)`
In `DeltaPML,PLgtPM`
`rArrPLgt(1)/(2)PP_(1)` (1)
Similarly, `LQgt(1)/(2)QQ_(1)` (2)
Adding (1) and (2), we get
`PL+LQgt(1)/(2)(PP_(1)+Q Q_(1))`
`rArrPQgt(1)/(2)(PP_(1)+Q Q_(1))`
So, PQ is more than mean of `PP_(1)andQ Q_(1)`.
So, `PQgemin(PP_(1),Q Q_(1))`
Let min `(PP_(1),Q Q_(1))=PP_(1)`.
Then `PQ^(2)gePP_(1)^(2)`
`=(x_(1)^(2)+1-x_(1))^(2)+(x_(1)^(2)+1-x_(1))^(2)`
`=2(x_(1)^(2)+1-x_(1))^(2)=f(x_(1))`
`rArrf'(x_(1))=4(x_(1)^(2)+1-x_(1))(2x_(1)-1)`
`=4((x_(1)-(1)/(2))^(2)+(3)/(4))(2x_(1)-1)`
Therefore, `f'(x_(1))=0` when `x_(1)=(1)/(2)`.
Also, `f'(x_(1))lt0" if "x_(1)lt(1)/(2)`
`andf'(x_(1))gt0" if "x_(1)gt(1)/(2)`
Thus, `f(x_(1))` is minimum when `x_(1)=(1)/(2)`.
Thus, at `x_(1)=(1)/(2)` point P is `P_(0)" on "C_(1)`.
`P_(0)-=((1)/(2),((1)/(2))^(2)+1)-=((1)/(2),(5)/(2))`
Similarly, `Q_(0)" on "C_(2)` will be image of `P_(0)` with respect to y=x.
`:.Q-=((5)/(4),(1)/(2))`