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

Question

Genesis Allen · Accepted Answer

#x=1/10(x-41)^2+211/2#

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)# and #(x_2,y_2)# is given by #sqrt((x_2-x_1)^2+(y_2-y_1)^2)# and distance of a point #(x_1,y_1)# from a line #ax+by+c=0# is #|(ax_1+by_1+c)/sqrt(a^2+b^2)|#

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

#|(x-103)/sqrt(1^2+0^2)|=|(x-103)/1|=|x-103|#

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

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

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

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

or #108^2+x^2-216x+41^2+y^2-82y=x^2+103^2-206x#

or #11664+x^2-216x+1681+y^2-82y=x^2+10609-206x#

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

or #10x=y^2-82y+2736#

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

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

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

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]}

Hazel Anglin · Answer

undefined The standard form of the equation of a parabola with a directrix at $ x = a $ and a focus at $ (h, k) $ is:

$$ (x - h)^2 = 4p(y - k) $$

where $ p $ is the distance between the vertex and the focus, and it's also the distance between the vertex and the directrix.

Given the focus at $ (108,41) $ and the directrix at $ x = 103 $, the vertex is halfway between them, so the vertex is at $ (103, 41) $. The distance between the focus and the directrix is $ |108 - 103| = 5 $, so $ p = 5 $.

Substitute the values into the standard form:

$$ (x - 103)^2 = 4 \cdot 5(y - 41) $$

$$ (x - 103)^2 = 20(y - 41) $$

So, the standard form of the equation of the parabola is $ (x - 103)^2 = 20(y - 41) $.