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The tangents to the curve y = (x - 2)^(2) - 1 at its points of intersectio with the line x - y = 3, intersect at the point |
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Answer» `((5)/(2), 1)` `y = (x - 2)^(2) - 1` `implies y = x^(2) - 4X + 3` Now, let `(x_(1) y_(1))` be the print of intersection of TANGENTS of parabola (i) and line x - y = 3, then Equation of CHORD of contact of point `(x_(1), y_(1))` w.r.t parabola (i) is `T = 0` `implies (1)/(2) (y + y_(1) = xx_(1) - 2 (x + x_(1)) + 3` `implies y + y_(2) = 2x (x_(1) - 2) - 4x_(1) + 6` `implies 2x (x_(1) - 2) - y = 4 x_(1) + y_(1) - 6`, this equation represent the line x - y = 3 only, so on comparing, we get `*(2(x_(1) - 2))/(1) = (-1)/(-1) = (4x_(1) + y_(1) - 6) ` `implies x_(1) = (5)/(2)` and `y_(1) = 1` So, the REQUIRED point is `((5)/(2), - 1)` |
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