1.

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

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


Discussion

No Comment Found

Related InterviewSolutions