The focus of parabola is (1,5) and its directrix is x + y + 2 = 0. Find the equation of the parabola, its vertex and length of latus rectum.

let F (1,5) be the focus of the parabola and the equation of the directrix is x+y+2=0...........(1)

let P(h.k) be any point on the parabola.
the distance from the focus to the point P = the distance from the directrix to the point P
PM=PF
(h-1)2+(k-5)2=h*1+k*1+21+1(h-1)2+(k-5)2=(h+k+2)222*[h2+1-2h+k2+25-10k]=h2+k2+4+2hk+4k+4h2h2+2k2-4h-20k+52=h2+k2+4+2hk+4k+4hh2+k2-8h-24k-2hk+48=0
put h = x and k = y
x2+y2-8x-24y-2xy+48=0 is the required equation of the parabola.

the axis of parabola passes through the focus and perpendicular to the directrix.
the slope of axis = -(-1)=1 [since slope of directrix is -1]
the equation of axis is
y-5=1*(x-1)y-5=x-1x-y+4=0 ...........(2)
the point of intersection of axis and directrix is intersection point of eq(1) and eq(2)
x+y+2=0x-y+4=02x+6=0x=-3-3+y+2=0y=1
the coordinates of k are (-3,1).
vertex is the mid-point of focus and point K
i.e. coordinates of vertex are 1-32,5+12i.e. (-1,3)
now distance from vertex to focus = (-1-1)2+(3-5)2=4+4=22
the length of the latusrectum = 4*22=82

hope this helps you

 

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