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The line `y = x + 1`is a tangent to the curve `y^2=4x`at the point(A) `(1, 2)` (B)`(2, 1)` (C) `(1, 2)` (D) `( 1, 2)`A. `(1, 2)`B. `(2, 1)`C. `(1, -2)`D. `(-1, 2)` |
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Answer» Correct Answer - A Equation of curve `y^(2) = 4x` ….(1) ` rArr 2y(dy)/(dx) = 4 rArr (dy)/(dx) = 2/y` Slope of tangent at point `(x, y) = 2/y` Equation of given tangent ` y = x + 1` Its slope = 1 ` :. 2/y = 1 rArr y= 2` put y = 2 in equation (1) `2^(2) = 4x rArr x = 1` ` :.` Point of contact = (1, 2) |
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