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Find the equation of the tangent line to the curve `y=x^2-2x+7`which is(a) parallel to the line `2x y + 9 = 0`(b) perpendicular to the line `5y 15 x = 13`. |
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Answer» Equation of curve ` y=x^(2) - 2x+7` …(1) ` rArr (dy)/(dx) = 2x -2` (a) Line 2x-y +9= 0 `rArr y = 2x +9` Its slope m=2 Slope of parallel line = 2 `:. 2x - 2 = 2` `rArr 2x - 2 = 2` ` rArr 2x = 4 ` ` rArr x = 2 ` put x = 2 in equation (1) ` y= 2^(2)-2 xx 2+7=7` and equation of tangent at point (2, 7) ` y-7=2(x-2)` ` rArr y- 7 = 2x - 4` `rArr 2x - y + 3 = 0` (b) Equation of line `5y - 15x = 13` ` rArr y = 3x + (13)/5` Its slope = 3 Slope of perpendicular line ` = - 1/3` `2x - 2 =- 1/3` ` rArr 2x = - 1/3 + 2 = 5/3 ` ` rArr x = 5/6` put ` x= 5/6` in equation (1) ` y=(5/6)^(2)-2xx5/6 + 7 = (25-60+252)/(36) = (217)/36` and equation of tangent at point `(5/6,(217)/36)` ` y- (217)/36 = - 1/3 (x-5/6)` (36y- 217)/36 =-((6x-5)/18)` ` rArr 12x+ 36y-227 = 0` |
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