Locus and Parabloa question. (1 Viewer)

[swifty13]

This isn't a very hard question but I was revising and got a bit confused with it.

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

Any help would be much appreciated

 

[stampede]

find focus of that parabola

use the co-ordinates of the focus + the point given (4, 8) to find the equation of the line (focal chord)

 

[Aesytic]

that parabola is a sideways version of the parabola x^2=4ay, with the x and y axes interchanged, so therefore your focus is also switched so that it's S(a,0) rather than (0,a)

to find the focal length, just do the same as you would with a normal parabola, except this time the coefficient of x is the value of 4a
4a=16
.'.a=4

Finding the vertex of the parabola is the same deal, and with this one it's the origin (0,0). Because the coefficient in front of the y^2 is positive, the curve "points" to the right, and therefore your focus would be S(4,0).
To find the equation of the focal chord that goes through the focus and (4,8), the x values of both points are 4, which means that the chord must be the vertical line x=4.