Question

the equation of the parabola shown can be written in the form

y^2=px or x^2=4py

if 4p= -68 then the length of the latus rectum is?

the endpoints of the latus rectum are?

please show me how to do this problem.i tried and i havent figured it out right

Answer 1

Latus Rectum The line segment through a focus of a conic section, perpendicular to the major axis, which has both endpoints on the curve.

length of latus recturm of parabola of form y^2 = px is p.

so if 4p = -68, so p = -17. so length of latus rectum is 17

length of latus recturm of parabola of form x^2 = 4py is 4p

if 4p = -68, so p = -17. so length of latus rectum is 68

so for parabola 1: y^2 = px, focus is (p/4,0)

Remember the latus rectum passes through focus, so the x-value of end points of the latus rectum would be same as the x-value of focus.

y^2 = p*p/4

y = +p/2 and -p/2

so end points = (p/4,-p/2) (p/4,p/2)

so for parabola 2: x^2 = 4py, focus is (0,p)

Remember the latus rectum passes through focus, so the y-value of end points of the latus rectum would be same as the y-value of focus.

x^2 = 4p*p

y = +2p and -2p

so end points = (-2p,p) (2p,p)