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y=x is tangent to the parabola `y=ax^(2)+c`. If a=2, then the value of c isA. 1B. `-1//2`C. `1//2`D. `1//8` |
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Answer» Correct Answer - 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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