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P(2, 2) is a point on the parabola y^(2)...

P(2, 2) is a point on the parabola `y^(2)=2x` and A is its vertex. Q is anoter point on the parabola such that PQ is perpendicular to AP. What is the length of PQ.

A

`sqrt2`

B

`2sqrt2`

C

`4sqrt2`

D

`6sqrt2`

Text Solution

Verified by Experts

The correct Answer is:
D
Equation of parabola is `y^(2)=2x`, so vertex lies at origin So, co-ordinates of vertex are A(0, 0).
Let `(x_(1), y_(1))` be the co-ordinates of the point Q
`:." "y_(1)^(2)=2x_(1)" ....(i)"`
`"and slope PQ"=(y_(1)-2)/(x_(1)-2)`
[co-ordinates of P is (2, 2) as givne ]
Also, slope of `AP=(2-0)/(2-0)=1`
Since, PQ and AP are perpendicular to each other, hence, slope of AP`xx` Slope of PQ=-1
So, `1xx((y_(1)-2)/(x_(1)-2))=-1`
`rArr" "y_(1)-2=-x_(1)+2`
`rArr" "x_(1)+y_(1)=4rArrx_(1)=4-y_(1)`
Putting value of `x_(1)` in equation (i)
`y_(1)^(2)=9-2y_(1)" or "y_(1)^(2)+2y_(1)-8=0`
`rArr" "y_(1)=-4" and "2`
Hence, co-ordinates of point Q are (8, -4).
So, required length `PQ=sqrt((8-2)^(2)+(-4-2)^(2))`
`=sqrt(36+36)=sqrt72=6sqrt2`
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