Point P on the curve, x = 2tan( y+ π/12), has a y-coordinate of π/4. Find an equation for the normal to the curve at P.

Step 1 - Firstly, we need to find the x-coordinate of point P. This will allow us to find the normal equation required in the question. To find the x-coordinate, we plug in the value of y given in the question, π/4, in the equation also given in the question. This gives us an x-coordinate of 2√3. Now we know point P has coordinates (2√3, π/4).Step 2 - Now we must differentiate the equation in the question to give dx/dy. This will help us find the gradient of the normal at P. To differentiate this, we must know that the derivative of tan(x) is just sec²(x), and thus, the derivative of 2tan(y+π/12) is 2sec²(y+π/12).Step 3 - Now we know that dx/dy = 2sec²(y+π/12), but we want to find dy/dx, as this form always give us the gradient, not dx/dy. To find dy/dx, we simply invert dx/dy by 'flipping it'. This means dy/dx = 0.5cos²(y+π/12).Step 4 - To find the gradient at P, we plug in the y value given in the question into our dy/dx. This gives us a value of 1/8.Step 5 - As the question asks for the gradient of the normal, and not the tangent, we must use the negative inverse of 1/8, which is -8. Examiners expect you to know that gradient of a normal to any tangent is just the negative inverse, so make sure you read the question carefully!Step 6 - Using our equation for a straight line, y-y₁ = m(x-x₁), we plug in our y-coordinate (given in question), x-coordinate (calculated in step 1) and gradient (calculated in step 5). This gives us the answer of:
y= -8x +16√3 + π/4