A line is drawn form `A(-2,0)` to intersect the curve `y^2=4x` at `Pa n dQ` in the first quadrant such that `1/(A P)+1/(A Q)<1/4dot` Then the slope of the line is always. (a) `>sqrt(3)` (b) `<1/(sqrt(3))` (c) `>sqrt(2)` (d) `>1/(sqrt(3))`

Options

A. `gtsqrt(3)`
B. `lt1//sqrt(3)`
C. `gtsqrt(2)`
D. `gt1//sqrt(3)`

Answer

A

Explanation

(1) Let `(-2+rcostheta,rsintheta)` lie on the parabola. Then,

`r^(2)sin^(2)theta-4(-2+rcostheta)=0`
`orr_(1)+r_(2)=(4costheta)/(sin^(2)theta),r_(1)r_(2)=(8)/(sin^(2)theta)`
Now, `(1)/(AP)+(1)/(AQ)=(r_(1)+r_(2))/(r_(1)r_(2))=(costheta)/(2)`
Given that `(1)/(AP)+(1)/(AQ)lt(1)/(4)`
`or costhetalt(1)/(2)`
`ortanthetagtsqrt(3)" "[becausecostheta" is decreasing and "tan theta" is increasing in "(0,pi//2)]`
`:.mgtsqrt(3)`