The distance between the origin and the tangent to the curve `y=e^(2x)+x^2` drawn at the point `x=0` is `(1,1/3)` (b) `(1/3,1)` `(2,-(28)/3)` (d) none of these

Options

A. ` (1)/(sqrt(5))`
B. ` (2)/(sqrt(5))`
C. ` (-1)/(sqrt(5))`
D. ` (2)/(sqrt(3))`

Answer

A

Explanation

Putting `x=0 ` in the given curve, we obtain ` y=1.`
So, the given point is `(0,1).`
Now, `y=e^(2x)+x^(2) rArr (dy)/(dx) = 2e^(2x)+ 2x rArr ((dy)/(dx))_((0","1))=2 `
The equation of the tangent at `(0,1)` is
` y-1=2(x-0) rArr 2x-y+1=0 " "...(i) `
` rArr " Required distance " = " Length of the " bot " from " (0,0) " on " (i) `
` rArr " Required distance " = (1)/(sqrt(5))`