Question

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

Answer 1

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

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)` 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]}