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| 1. |
The ordinates of the feet of three normals to the parabola `y^2=4ax` from the point (6a, 0) are |
| Answer» We need to draw three normals from the point `(6a,0)` on the parabola `y^2 = 4ax``Eqn.` of normal in parametric form:`y = -tx + 2at + at^3`This will pass from `(6a,0)`, so it will satisfy the above equation.`rArr 0 = -t*6a + 2at + at^3``rArr at^3 - 4at = 0``rArr at(t^2 - 4) = 0``rArr t = 0 or t = pm 2``rArr t_1 = 0, t_2 = 2, t_3 = -2`Normal point : `P(at^2, 2at)`From obtained values of t we can get three normal points:`A(0,0) , B(4a,4a) , C(4a,-4a)`So oordinates of the normals will be `0, 4a, -4a` | |