How do you derive eqn of tangent for this? (1 Viewer)

[enigma_1]

P(at^2, 2at) on the parabola y^2=4ax.

How do I show the tangent at P has equation x=ty-at^2 WITHOUT the use of implicit differentiation?
I did dx/dy and got y/2a, then I got the reciprocal of that which is 2a/y. When I used the point gradient formula inserting 2a/y as the tangent, why didn't it come out?

 

[QZP]

You forgot to sub point P into the dy/dx = 2a/y. How do you expect to get the tangent at P without knowing the gradient there?

 

[Chris100]

I don't know if i used the right method, i just reversed everything and I got it

Can someone clarify why i have to reverse everything for me to get the right answer?

 

[enigma_1]

LOL I see what you mean QZP I was freaked out by the questions that I didn't even end up doing that hahahah!!

Ohk I see what you did there, Chris.
I just tried what QZP said and I got it.
You know how you did dx/dy = t?
After that, ,just do dy/dx = 1/t and then just sub that into the point gradient formula. I reckon use this method to be safe. I'm not sure if you can reverse it like that actually.

 

[RealiseNothing]

There's two ways without using implicit differentiation:

1)



Then just differentiate this.

2)











Woops just realised I got my and mixed around, just reverse them and it's the same thing.