Find the equation of the parabola whose focus is `S(-1,1)` and directrix is `4x+3y-24=0` .

Answer

`9x^(2)+16y^(2)-24xy+24x+94y-526=0`

Explanation

Let P(x,y) be any point on the parabola. Since the distance of P from the focus is equal to its distance from the directrix, we have
`PS=PQorPS^(2)=PQ^(2)`
`or" "(x+1)^(2)+(y-1)^(2)=[((4x+3y-24))/(5)]^(2)`
`i.e.," "9x^(2)+16y^(2)-24xy+24xy+242x+94y-526=0`
This is the required equation of the parabola.
The axis is a line through S(-1,1) and perpendicular to the directrix 4x+3y-24=0. Thus, the equation of the axis is
`3(x+1)-4(y-1)=0`
`or3x-4y+7=0`
The axis and the directrix intersect at B.
Solving them, we get B (3,4).
The vertex A is the midpoint of S(-1,1) and B (3,4).
Thus, `A-=(1,(5)/(2))`
Also, `AS=sqrt(2^(2)+((3)/(2))^(2))=(5)/(2)`
Hence, the latus rectum is a straight line through the focus S and parallel to the directrix.
Hence, its equation is 4x+3y+1=0.