The tangent of the angle beta is also the gradient of the upsloping line (dy/dx) since tangent is opposite (vertical distance) divided by adjacent (horizontal distance).

Then the perpendicular line has gradient -1/tangent since product of gradient of two perpendicular lines is -1.

This diagram is only one of the possibilities.
OP isn't always going to be the same as the tangent to the point at P.

Rachel, you posted this yesterday and I gave 2 comments. Did you see them? If yes, was there any part you didn't understand?

Hi! I didnt see any responses and so i decided to repost this question

It was under the comments section... Nvm I will try to explain again

Notice that the normal to the curve at P forms a right angled triangle with the
x axis and perpendicular line from P to the axis.

The angle α is the angle of interest.

We can use trigo to find the gradient of the normal.


Let the foot of the perpendicular from P to the x-axis be J.
Let the point of intersection of the normal to P and the x axis be K.

Now the slope of the normal is just :

Vertical height of PJ ÷ JK

And this actually equals tan α since

tan α = opp/adj of △PJK (with α as the target angle) = PJ/PK


Since the normal and the tangent to P are perpendicular to each other, the product of their gradients is -1. And the gradient of tangent to P is just dy/dx at point P.

So tan α × dy/dx = -1

And therefore

tan α = -1/ dy/dx

In your answer scheme, the α is obtuse. But it can actually be acute or even 90° as well.

In any of these cases, trigo can be used to find dy/dx.


The next part is just trigo manipulation using the property

tan(180° - θ) = - tan θ (degrees)
Or
tan(π - θ) = - tan θ (radians)

so since α + 2β = π,

then tan α = tan (π - 2β) = -tan(2β)

The trigo identity tan2θ = 2tanθ/(1 - tan²θ)
is applied.

This is done so that tanβ = y/x can be introduced to get to the desired

Sorry for the trouble! Thank you for ur detailed explanation, i understand it clearly now! :)

Welcome, no worries