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PQ is a normal chord of the parabola y^2...

`PQ` is a normal chord of the parabola `y^2= 4ax` at `P,A` being the vertex of the parabola. Through P a line is drawn parallel to `AQ` meeting the x-axis in R. Then the length of `AR` is : (A) equal to the length of the latus rectum (B) equal to the focal distance of the point P (C) equal to the twice of the focal distance of the point P (D) equal to the distance of the point P from the directrix.

A

equal to the length of the latus rectum

B

equal to the focal distance of the point P

C

equal to twice focal distance of the point P

D

equal to the distance of the point P from the directrix

Text Solution

Verified by Experts

The correct Answer is:
C

(3) `t_(2)=-t_(1)-(2)/(t_(1))ort_(1)t_(2)=-t_(1)^(2)-2`
The equation of the line through P parallel to AQ is
`y-2at_(1)=(2)/(t_(2))(x-at_(1)^(2))`
Put y=0. Then,
`x=at_(1)^(2)-at_(1)t_(2)`
`=at_(1)^(2)-a(-2-t_(1)^(2))`
`=2a+2at_(1)^(2)`
`=2a+2at_(1)^(2)`
`=2a+2at_(1)^(2)`
`=2(a+at_(1)^(2))`
=twice the focal distance of P
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