Question

Find the equation of parabola whise focus is (2,5) and directrix is 3x+4y+1=0???? ( please write the answer ona paper and send the answer).

Answer 1

Parabola :   F(2,5)
Directrix D  = 3x + 4y + 1 = 0
Let a point P(h,k) be on the Parabola. 

By definition, parabola is the locus of a point P whose ⊥ distance from the directrix D is same as the distance from Focus F.

  PF² = (h-2)² + (k-5)²
  (Distance of P from D)²  = (3 h + 4k + 1)² / (3²+4²)

 Hence, [h²-4h+4+ k²-10k+25] *25 = 9h²+16k²+1+24hk+8k+6h
             16h² + 9k² -106 h -258 k - 24 h k + 724 = 0
   Replace (h,k) by (x,y)

Parabola:   16 x² + 9 y² - 106 x - 258 y - 24 x y + 724 = 0

Rewrite it as:
   (4x - 3y + w)² = w²+ 8w x - 6 wy + 106 x + 258 y - 724
                         = x(8w+106) + y(258-6w) + (w²-724)
  Finding w from  (258 - 6w)/(8w+106) = 4/3,  we get: w = 7

  So equation of parabola :  (4x- 3y +7)² = 27 (6x + 8y - 25)
            Vertex at : (19/50, 71/25).     Axis:  6x+8y - 25 = 0

Answer 2

given ,
focus (2 ,5)
and directrix 3x + 4y + 1 =0

we know ,
a parobala is the locus of a point which moves in a plane such that its distance from a fixed point is equal to its distance from fixed line. (directrix)

e.g let a point ( r , s )
then by concept of parabola ,
| 3r + 4s + 1|/5 = {(r -2)^2 + (s -5)^2}^1/2
take square both side ,
(3r + 4s + 1)^2/25 = (r -2)^2 + (s -5)^2

9r^2 + 16s^2 + 1 +24rs + 8s +6r = 25r ^2 +25s^2 +625 + 100 -100r - 250s

16r ^2 + 9s^2 -24rs -106r -258s +724 =0

now ,
put. r = x. and s = y

16x ^2 + 9y^2 -24xy -106x -258y +724 =0