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At a point P on the parabola y^(2) = 4ax...

At a point P on the parabola `y^(2) = 4ax`, tangent and normal are drawn. Tangent intersects the x-axis at Q and normal intersects the curve at R such that chord PQ subtends an angle of `90^(@)` at its vertex. Then

A

`PQ = 2a sqrt(6)`

B

`PR = 6a sqrt(3)`

C

area of `Delta PQR = 18 sqrt(2)a^(2)`

D

`PQ = 3a sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
Let `P -= (at_(1)^(2),2at_(1))`
Equation of tangent at P is `t_(1)y = x + at_(1)^(2)`. On x-axis, `y =0`, so `Q -= (-at_(1)^(2),0)`.
Normal at P intersects the parabola at `R (at_(2)^(2), 2at_(2))`.
`rArr t_(2) =- t_(1) -(2)/(t_(1))` (i)
Slope of `OP, m_(1) = (2)/(t_(1))`
Slope of `OR, m_(2) = (2)/(t_(2))`
`:' OP _|_ OR rArr t_(1)t_(2) =- 4`
From (i), `t_(1) = +- sqrt(2)`
`:. t_(2) = +- 2sqrt(2)`
`PQ = 2at_(1) sqrt(t_(1)^(2)+1) = 2a sqrt(6)`
Area of `DeltaPQR = (1)/(2) |{:(2a,2sqrt(2)a,1),(-2a,0,1),(8a,-4sqrt(2)a,1):}| =18 sqrt(2)a^(2)`
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