Locus and Parabola Question (1 Viewer)

[kloudsurfer]

Hey,

I need some help with this question:

'Find the equation of the focal chord that cuts the curve y^2 = 16x at (4,8)

I know how to do it, and I know the answer, but I cant seem to get it with my working out.

So i found the focus (4,0), and then you have to find the eqution of the line.

The gradient is 0 (and ive already figured out that the equation is x=4 but anyway)

So when Isub the points into y-y1 = m(x-x1) I get either

y=8 (sub in (4,8)) or y = 0 (sub in (4,0))

But how do I actually get the equation x=4?

Thanks

 

[Forbidden.]

tried the y-y₁/x-x₁=y₂-y₁/x₂-x₁

formula ?

i wish i could to this on paper

 

[Riviet]

The gradient is 0 (and ive already figured out that the equation is x=4 but anyway)

But x=4 is a vertical line on the number plane and has an undefined gradient. Therefore you can't use the gradient formula to find the equation of the line. You simply need to say that since it cuts through (4,0) and (4,8), then it must be the line x=4.

 

[pLuvia]

The line x=4 is the lactus rectum a type of focal chord, not necessarily the focal chord for this answer.

Since you know the focus is (4,0) and it has to pass through (4,8), then it's obvious that the line is x=4.

And also note x=4 does not have a gradient of zero, a line y=1,2,3 etc has a gradient of zero

 

[kloudsurfer]

But x=4 is a vertical line on the number plane and has an undefined gradient. Therefore you can't use the gradient formula to find the equation of the line. You simply need to say that since it cuts through (4,0) and (4,8), then it must be the line x=4.

Fair enough. I just thought I was doing something wrong because i didnt actually get that answer when i worked it out. But it makes sense now.

Thanks everyone