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Tangents PA and PB are drawn to parabola...

Tangents `PA` and `PB` are drawn to parabola `y^(2)=4x` from any arbitrary point `P` on the line `x+y=1`.
Then vertex of locus of midpoint of chord `AB` is

A

`(-1,-2)`

B

`(3/2,1)`

C

`(-3/2,-1)`

D

`(3/2,-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Let point of line is `(alpha,1-alpha)`
So, equation of chord of contact is
`y(1-alpha)=2(x+alpha)`
`2x-y(1-alpha)+2alpha=0` …………..(1)
and let `R(h,k)` is midpoint of chord then equation of chord `T=S_(1)`
`ky-2(x+y)=k^(2)-4h`
`2x-ky+k^(2)-2h=0`........(2)
On comparing (1) and (2)
`1-alpha=k,2alpha=k^(2)-2h`
`implies2(1-k)=k^(2)-2h`
`y^(2)+2y=2(x+1)`
`(y+1)^(2)=2(x+1)+1`
`(y+1)^(2)=2(x+3/2)`
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