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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 theseA. `sqrt(2)`B. `2sqrt(2)`C. `4sqrt(2)`D. none of these |
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Answer» Correct Answer - C (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)` |
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