How do you find an equation of a parabola given focus (10,1) and directrix x=5?

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Answer 1 · 2017-02-26T15:04:52

# (y-1)^2=(x-5)^2-(x-10)^2=5(2x-15), or, #

# y^2-2y-10x+76=0.#

We use the following Focus-Directrix Property (FDP) of

Parabola to solve this Problem :

FDP : Let a pt. #S# and a line #d,# be the focus & directrix of a

parabola, resp. If #P# is any pt. on the parabola, then, #P# is

equidistant from #S# and #d.#

Focus is #S=S(10,1)" and the eqn. of dir. is "d : x-5=0.#

Let #P(x,y)# be any pt. on the parabola.

#:." the Dist. SP="sqrt{(x-10)^2+(y-1)^2}......................(1).#

The #bot-dist." from "P" to "d" is ="|x-5|/sqrt{1^2+0^2}=|x-5|...(2)#

FDP#rArrsqrt{(x-10)^2+(y-1)^2}=|x-5|.#

#rArr (y-1)^2=(x-5)^2-(x-10)^2=5(2x-15), or, #

# y^2-2y-10x+76=0.#

Enjoy Maths.!