Normals at two points `(x_1y_1)a n d(x_2, y_2)` of the parabola `y^2=4x` meet again on the parabola, where `x_1+x_2=4.` Then `|y_1+y_2|` is equal to `sqrt(2)` (b) `2sqrt(2)` (c) `4sqrt(2)` (d) none of these

Options

A. `sqrt(2)`
B. `2sqrt(2)`
C. `4sqrt(2)`
D. none of these

Answer

C

Explanation

(3) Normal at point `P(x_(1),y_(1))-=(at_(1)^(2),2at_(1))` meets the parabola at R `(at^(2),2at)`. So,
`t=-t_(1)-(2)/(t_(1))` (1)
Normal at point `P(x_(2),y_(2))-=(at_(2)^(2),2at_(2))` meets the parabola at R `(at^(2),2at)`. So,
`t=-t_(2)-(2)/(t_(2))` (2)
From (1) and (2), we get
`-t_(1)-(2)/(t_(1))=-t_(2)-(2)/(t_(2))`
`:.t_(1)t_(2)=2`
Now given that `x_(1)+x_(2)=4`. Therefore,
`t_(1)^(2)+t_(2)^(2)=4`
`or(t_(1)+t_(2)^(2))=4+4=8`
`or|t_(1)+t_(2)|=2sqrt(2)`
`or|y_(1)+y_(2)|=4sqrt(2)`