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Question Number 5113 by FilupSmith last updated on 14/Apr/16
How do you find the: (1) Focal point (2) Directrix for y=ax^2 +bx+c For simplicity, lets assume it goes throigh point (0, 0).
$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{find}\:\mathrm{the}: \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{Focal}\:\mathrm{point} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{Directrix} \\ $$$$\mathrm{for}\:{y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{simplicity},\:\mathrm{lets}\:\mathrm{assume}\:\mathrm{it}\:\mathrm{goes}\:\mathrm{throigh} \\ $$$$\mathrm{point}\:\left(\mathrm{0},\:\mathrm{0}\right). \\ $$
Answered by Yozzii last updated on 14/Apr/16
For a parabola of the form 4py=x^2 , its focus is F(0,p) and its directrix is given by y=−p. We can rewrite y=ax^2 +bx+c as y=a(x+(b/(2a)))^2 +c−(b^2 /(4a)) or (1/a)(y+(b^2 /(4a))−c)=(x+(b/(2a)))^2 . Let J=y+(b^2 /(4a))−c, N=x+(b/(2a)). (1/a)J=N^2 . J and N represent the coordinate system (N,J). ⇒4p=(1/a)⇒p=(1/(4a)). So, in (N,J) coordinates F(0,(1/(4a))) and directrix J=−(1/(4a)). Since N=x+(b/(2a)) and J=y+(b^2 /(4a))−c we can convert back to xy−coordinates giving F(−(b/(2a)),((1−b^2 )/(4a))+c) and directrix y=c−((1+b^2 )/(4a)).
$${For}\:{a}\:{parabola}\:{of}\:{the}\:{form}\:\mathrm{4}{py}={x}^{\mathrm{2}} ,\: \\ $$$${its}\:{focus}\:{is}\:{F}\left(\mathrm{0},{p}\right)\:{and}\:{its}\:{directrix}\:{is} \\ $$$${given}\:{by}\:{y}=−{p}.\: \\ $$$${We}\:{can}\:{rewrite}\:{y}={ax}^{\mathrm{2}} +{bx}+{c}\:{as} \\ $$$${y}={a}\left({x}+\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{2}} +{c}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}} \\ $$$${or}\:\frac{\mathrm{1}}{{a}}\left({y}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}−{c}\right)=\left({x}+\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{2}} . \\ $$$${Let}\:{J}={y}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}−{c},\:{N}={x}+\frac{{b}}{\mathrm{2}{a}}. \\ $$$$\frac{\mathrm{1}}{{a}}{J}={N}^{\mathrm{2}} .\:{J}\:{and}\:{N}\:{represent}\:{the}\: \\ $$$${coordinate}\:{system}\:\left({N},{J}\right).\: \\ $$$$\Rightarrow\mathrm{4}{p}=\frac{\mathrm{1}}{{a}}\Rightarrow{p}=\frac{\mathrm{1}}{\mathrm{4}{a}}.\:{So},\:{in}\:\left({N},{J}\right)\:{coordinates} \\ $$$${F}\left(\mathrm{0},\frac{\mathrm{1}}{\mathrm{4}{a}}\right)\:{and}\:{directrix}\:{J}=−\frac{\mathrm{1}}{\mathrm{4}{a}}.\:{Since}\:{N}={x}+\frac{{b}}{\mathrm{2}{a}} \\ $$$${and}\:{J}={y}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}−{c}\:{we}\:{can}\:{convert}\:{back} \\ $$$${to}\:{xy}−{coordinates}\:{giving}\: \\ $$$${F}\left(−\frac{{b}}{\mathrm{2}{a}},\frac{\mathrm{1}−{b}^{\mathrm{2}} }{\mathrm{4}{a}}+{c}\right)\:{and}\:{directrix}\:{y}={c}−\frac{\mathrm{1}+{b}^{\mathrm{2}} }{\mathrm{4}{a}}. \\ $$$$ \\ $$$$ \\ $$

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