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The tangent to `y=a x^2+b x+7/2a t(1,2)`is parallel to thenormal at the point `(-2,2)`on the curve `y=x^2+6x+10 ,`then`a=-1`b. `a=1`c. `b=5/2`d. `b=-5/2`A. `a = 1`B. `a = -1`C. `b = (-15)/(2)`D. `b = (5)/(2)` |
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Answer» Correct Answer - A `y = x^(2) + 6x + 10` `:. (dy)/(dx) = 2x + 6` `(dy)/(dx)|(-2,2) = 2` `:.` slope of Normal,`m = - 1//2` `:. y = ax^(2) + bx + (7)/(2)` Passes through `(1,2) rArr a+ b = -3//2`…..(1) Also `(dy)/(dx)|(1,2) = 2a + b` `rArr - (1)/(2) = 2a + b`......(2) `[1]` and [2] `rArr a = 1 , b = -5//2` |
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