How do you find an equation of a parabola given vertex (7,0) and focus (0,0)?

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Answer 1 · 2017-12-19T09:32:35

#y^2=-28x+196#

As the vertex is #(7,0)# and focus is #(0,0)#, the line joining them is axis of symmetry i.e. #y=0#

Directrix is perpendicular to axis of symmetry and hence its equation is of the type #x=k#

Further as vertex is equidistant from focus and directrix and is also between the two, we must have directrix as #x=14#

Now parabola is locus of a point, say #(x,y)#, which moves so that its distance from focus and directrix is always equal,

as distance of #(x,y)# from #(0,0)# is #sqrt(x^2+y^2)#

and distance of #(x,y)# from #x=14# is #|x-14|#,

equation of parabola is #sqrt(x^2+y^2)=|x-14|#

or #x^2+y^2=x^2-28x+196#

or #y^2=-28x+196#

graph{(y^2+28x-196)(x^2+y^2-0.2)((x-7)^2+y^2-0.2)(x-14)=0 [-33.4, 46.6, -20.16, 19.84]}