Question

How do you identify the vertex, focus, directrix and the length of the latus rectum and graph `x=3y^2+4y+1`?

Answer 1

Focus = (-0.6667,-0.25
Vertex = (-0.6667, -0.3333)
Directrix = 0.4167
Length of latus rectum = 5.8214

Explanation:

Given Eqn of parable `x=3y^2+4x+1`
It is in the form `ax^2+bx+c`
`a=3, b=4, c=1`

Focus `=((-b)/2a, c-((b^2-1)/(4a))`
`(-b)/(2a)=-4/6=-(2/3)`
`c-(b^2-1)/(4a)=1-(15/12)=-(1/4)`
Focus `(-2/3,-1/4)~~(-0.6667,-0.25)`

Vertex `=-b/(2a),-(b^2-4ac)/(4a)`
`-b/(2a)=-(2/3)`
`-(b^2-4ac)/(2a)=-(16-12)/12=-(1/3)`
Vertex `(-2/3,-1/3)~~(-0.6667,-0.3333)`

Directrix `x=>c-(b^2+1)/(4a)`
`x=>1-(17)/12=-(5/12)~~0.4167`

Length of latus rectum of parabola is 4p where p is the distance from focus to vertex.
`p=sqrt((16/9)+(49/144))=sqrt(305/144)`
`4p=(4/12)*sqrt305~~5.8214`#