The tangent at any point `(x , y)` of a curve makes an angle `tan^(-1)(2x+3y)` with x-axis. Find the equation of the curve if it passes through (1,2).

Options

A. ` 6x+9y +2 = 26 e^(3(x-1))`
B. ` 6x- 9y +2 = 26 e^(3(x-1))`
C. ` 6x + 9y -2 = 26 e^(3(x-1))`
D. ` 6x - 9y -2 = 26 e^(3(x-1))`

Answer

a

Explanation

Given , `(dy)/(dx) = tan theta = 2x +3y`
Put ` 2x + 3y = z rArr 2 +3 (dy)/(dx) = (dz)/(dx)`
` rArr (dy)/(dx) = ((dz)/(dx) - 2 ) 1/3 `
` :. (dz)/(dx) - 2 = 3z rArr (dz)/(3z + 2) = dx`
On integrating , we get
`(log (3z+2))/3 = x+C`
` rArr (log (6x + 9y +2))/3 = x=C`
Since , it passes through (1,2)
` :. (log (6+18+2))/3 = 1+C rArr C = (log 26)/3 -1`
` :. (log (6x +9y+2))/3 = x + (log 26)/3 -1`
` rArr log ((6x+9y+2)/26) = 3 ( x-1)`
` rArr 6x + 9y +2 = 26 e^(3(x-1))`