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y=x is tangent to the parabola y=ax^(2)+...

y=x is tangent to the parabola `y=ax^(2)+c`.
If a=2, then the value of c is

A

1

B

`-1//2`

C

`1//2`

D

`1//8`

Text Solution

Verified by Experts

The correct Answer is:
D
(4)
`y=ax^(2)+c`
`:.(dy)/(dx)=2ax=1`
Therefore, the point of contact of the tangent is
`((1)/(2a),(1)/(4a)+c)`
Since it lies on y=x, we have
`c=(1)/(4a)` Thus, c=1/8 for a=2.
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