Thread: NEED HELP WITH ALGEBRA PROOF! Tangent to a parabola y^2=4ax

---

Okay, can anyone help me with this problem- I will try to have a go myself as well.

Okay so there is a normal x and y axis. On this axis is a parabola y^2=4ax, flipped so that it looks like the letter C to the right of the y axis. The parabola is touching the y axis, with a vertex of (0,0). A line called x=a passes through the parabola so that it passes through the x axis and touches 2 opposite points of the parabola (this line is called a lactus rectum). The question asks to prove that 2 tangents (straight lines) that pass through the end points of the lactus rectum, intersects on the x axis at (-a,0). Ill post a diagram as well




Im unsure on how to go about this ??

Im guessing that we have to find the equations for the 2 tangents, and then make these two equations equal to eachother, because that would be the point where they intersect, then solve to find the point of intersection is at (-a,0)

Heres a diagram:

I think the end polnts of the lactus rectum are (a,y1) and (a,-y1), I think,,.... im not too sure

Take f(y) = then f'(y) = ,
Since the line is x = a, or or ,
so f'(-2a) =
f'(2a) =

so
and

so the above represent the tangent lines at point y = -2a, y =2a

so settting them equal to each other (Note f(-2a)=f(2a) since f is quadratic)
so we have , or y = 0 is where they intersect, inputting y = 0, in any of the tangent eqn, we get or so x = -a, y = 0, the tangent lines intersect.

Thank you so much , for replying . Umm im a bit confused about a few parts: so in the first step you re arranged the equation y^2=4ax and then differentiated it to find the gradient function did you go like : y^2=4ax => 2y. dy/dx=4ax^0 => 2y. dy/dx=4a => dy/dx= 4a/2y => dy/dx=2a/y because that's how I would have done it probably? Im confused how you got y/2a.

since i did so if i let x = G(y), and G(y) = , then i can do to get

i fixed the error i made, its good to go.

okay that make sense now , so basically that equation f(y)=y/2a is the equation for the slope of a tangent, at different points of (x) because x=f(y) im guessing- is this right????

. then you pick the x points (-a) and (a) and sub then into this function to find the tangent slopes at these points (im just confused on how the equation makes sense because wouldn't you find the equation of the tangent which passes through (a,y) and (a,-y) then make them equal?.

the change from y = f(x) to x = f(y) seems to be confusing you. Do what i did in the first post, where ever i put y, put x and see if it makes sense if our function was instead of

my last 2 questions are 1. how did you get the equation fot T1 and T2 (is y-y1=m(x-x1) where the points are (a,2a) slope =1 and (-a,2a) slope -1) so then would it be y-2a=1(x-a) and for the point (a,-2a) the equation y+2a=-1(x-a).... just dont get from this point onwards. It would be cool if you could exaplin this part to me using y=mx+c or y-y1=m(x-x1) step by step. Thanks

What is a derivative? A derivative at a point in your domain lets you approximate your function f(x) by a tangent line, the equation for this tangent line is f(x) = f(a) + f'(a)(x-a) . you can put . I only used T(x) to emphasize that it is merely an approximation to the function f(x) near point a, and doesnt equal f(x) everywhere on your domain.

where id you get the equation for the tangent line? is it basically y=mx+c?