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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the points on the curve \( y = 5x^2-2x^3\) at which the tangent is parallel to the line \( y-4x=5.\)

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Let x1,y1 be the points on the curve then, y1 = 5 x1^2 - 2 x1^3 Diffrentiating, the eqn of curve.. dy\dx = 10 x - 6 x^2 dy\dx on x1,y1 10 x1 - 8 x1^2 If tangent is parallel to y-4x = 5 Then, as we know it is the eqn of the straight line, General eqn of line is y= mx + c Where m = slope Therefore, comparing it wid givn eqn ol line, we get slope = 4 Therefore, as two slopes are equal we get 10 x1 - 6 x1^2 = 4 Or, 6 x1^2 - 10 x1 + 4 = 0 3 x1^2 - 3 x1 - 2 x1 + 2 = 0 3 x1 ( x1 - 1) - 2 ( x1 - 1) = 0 x1 = 2\3, 1 When x1 = 2\3, y1 = 44\27 When x1 = 1, y1 = 3. Hence, points on the curve are ( 2\3, 44\27), (1, 3)
answered Apr 3 by sarveshchauhan185
 
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