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The tangent to curve y_1=ax^2 + bx + 7/2 at (1,2) is parallel to normal at point (-2,2) on curve y_2=x^2 + 6x+10, then value of (a-2b) is |
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Answer» SOLUTION :(1,2) LIES on `y_1` `2=a+b+7/2 rArr a+b =-3/2`…(1) `((dy_1)/(dx))_(1,2)=-1/((dy_2)/(dx))_(2,2)` `2a+b=-1/2` From (1) & (2) a=1 & b=`-5/2` a-2b=6 |
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