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The area between the parabola `y^2 = 4x` , normal at one end of latusrectum and X-axis in sq.units isA. `60^(@)`B. less then `60^(@)`C. more then`60^(@)`D. less then `45^(@)` |
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Answer» Correct Answer - C Let `y+t_(1)x=2t_(1)+t_(1)^(3)` be a normal at `P(t_(1)^(2), 2t_(1))` to the parabola `y^(2)=4x`. Suppose it cuts the parabola again at `Q(t_(2)^(2), 2t_(2))`. Then, `t_(2)=-t_(1)-2/t_(1)` The equation of the normal at Q is `y+t_(2)x=2t_(2)+t_(2)^(3)` Suppose it makes an angle `theta` with the positive direction of x-axis. `:." "tantheta=-t_(2)` `rArr" "tantheta=t_(1)+2/t_(1)` `rArr" "|tantheta|=|t_(1)+2/t_(1)|ge2sqrt(t_(1)xx2/t_(1))" "["Using:"AMgeGM]` `rArr" "|tantheta|ge2sqrt2rArr0gt60^(@)` |
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