What is the standard form of the equation of the parabola with a directrix at x=103 and a focus at (108,41)?

1 Answer
Nov 17, 2017

x=1/10(x-41)^2+211/2x=110(x41)2+2112

Explanation:

A parabola is the locus of a point, which moves so that its distance from a given line called directrix and a given point called focus, is always equal.

Now, the distance between two pints (x_1,y_1)(x1,y1) and (x_2,y_2)(x2,y2) is given by sqrt((x_2-x_1)^2+(y_2-y_1)^2)(x2x1)2+(y2y1)2 and distance of a point (x_1,y_1)(x1,y1) from a line ax+by+c=0ax+by+c=0 is |(ax_1+by_1+c)/sqrt(a^2+b^2)|ax1+by1+ca2+b2

Coming to parabola with directrix x=103x=103 or x-103=0x103=0 and focus (108,41)(108,41), let the point equidistant from both be (x,y)(x,y). The distance of (x,y)(x,y) from x-103=0x103=0 is

|(x-103)/sqrt(1^2+0^2)|=|(x-103)/1|=|x-103|∣ ∣x10312+02∣ ∣=x1031=|x103|

and its distance from (108,41)(108,41) is

sqrt((108-x)^2+(41-y)^2)(108x)2+(41y)2

and as the two are equal, equation of parabola would be

(108-x)^2+(41-y)^2=(x-103)^2(108x)2+(41y)2=(x103)2

or 108^2+x^2-216x+41^2+y^2-82y=x^2+103^2-206x1082+x2216x+412+y282y=x2+1032206x

or 11664+x^2-216x+1681+y^2-82y=x^2+10609-206x11664+x2216x+1681+y282y=x2+10609206x

or y^2-82y-10x+2736=0y282y10x+2736=0

or 10x=y^2-82y+273610x=y282y+2736

or 10x=(y-41)^2+105510x=(y41)2+1055

or in vertex form x=1/10(x-41)^2+211/2x=110(x41)2+2112

and vertex is (105 1/2,41)(10512,41)

Its graph appears as shown below, along with focus and directrix.

graph{(y^2-82y-10x+2736)((108-x)^2+(41-y)^2-0.6)(x-103)=0 [51.6, 210.4, -13.3, 66.1]}