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A normal drawn to the parabola `=4a x`meets the curve again at `Q`such that the angle subtended by `P Q`at the vertex is `90^0dot`Then the coordinates of `P`can be`(8a ,4sqrt(2)a)`(b) `(8a ,4a)``(2a ,-2sqrt(2)a)`(d) `(2a ,2sqrt(2)a)`A. `(8a,4sqrt(2)a)`B. (8a,4a)C. `(2a,-2sqrt(2)a)`D. `(2a,2sqrt(2)a)` |
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Answer» Correct Answer - C::D 3,4 `t_(2)=-t_(1)-(2)/(t_(1))` Also, `(2at_(1))/(at_(1)^(2))xx(2at_(2))/(at_(2)^(2))=-1` `ort_(1)t_(2)=-4` `:.(-4)/(t_(1))=-t_(1)-(2)/(t_(1))` `ort_(1)^(2)+2=4andt_(1)=pmsqrt(2)` So, the point can be `(2a,pm2sqrt(2)a)`. |
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