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The tangent to the curve y = x^2 + 3x w...

The tangent to the curve `y = x^2 + 3x` will pass through the point `(0, - 9)` if it is drawn at the point

A

`x+ 25y =0`

B

`5x+y =0`

C

`5x-y =0`

D

`x -25y =0`

Text Solution

Verified by Experts

`y = (x+9)/(x+5) =1+ (4)/(x+5)`
`(dy)/(dx)` at `(x_(1),y_(1)) = (-4)/((x_(1)+5)^(2))`
`:.` Equation of tangent
`y - y_(1) = (-4)/((x_(1)+5)^(2)) (x-x_(1))`
`y -1 - (4)/(x_(1)+5) = (-4)/((x_(1)+5)^(2)) (x-x_(1))`
Since it passes through (0,0)
`(x_(1)+5)^(2) + 4(x_(1)+5) + 4x_(1) =0`
`x_(1) =- 15` or `x_(1) =-3`. So equations are `x + 25y = 0` or `x + y =0`.
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