Hope this helped. :)
Any point, (x0,y0)(x0,y0) on the parabola satisfies the definition of parabola, so there are two distances to calculate:
Distance between the point on the parabola to the focusDistance between the point on the parabola to the directrixTo find the equation of the parabola, equate these two expressions and solve for y0y0 .
Find the equation of the parabola in the example above.
Distance between the point (x0,y0)(x0,y0) and (a,b)(a,b) :
(x0−a)2+(y0−b)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√(x0−a)2+(y0−b)2
Distance between point (x0,y0)(x0,y0) and the line y=cy=c :
∣∣y0−c∣∣| y0−c |
(Here, the distance between the point and horizontal line is difference of their yy -coordinates.)
Equate the two expressions.
(x0−a)2+(y0−b)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=∣∣y0−c∣∣(x0−a)2+(y0−b)2=| y0−c |
Square both sides.
(x0−a)2+(y0−b)2=(y0−c)2(x0−a)2+(y0−b)2=(y0−c)2
Expand the expression in y0y0 on both sides and simplify.
(x0−a)2+b2−c2=2(b−c)y0(x0−a)2+b2−c2=2(b−c)y0
This equation in (x0,y0)(x0,y0) is true for all other values on the parabola and hence we can rewrite with (x,y)(x,y) .
Therefore, the equation of the parabola with focus (a,b)(a,b) and directrix y=cy=c is
(x−a)2+b2−c2=2(b−c)y