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how to find the equation of directrix if vertex is(1,-1)&focus is(2,3)
vertex(1,-1) , focus(2,3) (given) let directrix cuts the axis of parabola at (h,k) then vertex is midpoint of line joining focus & point at which directrix cuts the axis of parabola... from above statement , (h+2)/2 = 1 & (k+3)/2 = -1 (h,k) = (0,-5) slope of directrix is m & slope of axis is m1 then mm1 = -1 m1 = 4/1 = 4 m = -1/4 now eq of directrix : (y+5) = (x-0)(-1/4) x+4y+20 = 0 .................1 since eccentricity of parabola is 1 so SP = ePM distance of any point from directrix = e (distance of any point on parabola from focus) x+4y+20/(17)1/2 = [(x-2)2+(y-3)2]1/2 this is the required eq approve if u like my ans
vertex(1,-1) , focus(2,3) (given)
let directrix cuts the axis of parabola at (h,k) then
vertex is midpoint of line joining focus & point at which directrix cuts the axis of parabola...
from above statement , (h+2)/2 = 1 & (k+3)/2 = -1
(h,k) = (0,-5)
slope of directrix is m & slope of axis is m1 then
mm1 = -1
m1 = 4/1 = 4
m = -1/4
now eq of directrix : (y+5) = (x-0)(-1/4)
x+4y+20 = 0 .................1
since eccentricity of parabola is 1 so
SP = ePM
distance of any point from directrix = e (distance of any point on parabola from focus)
x+4y+20/(17)1/2 = [(x-2)2+(y-3)2]1/2
this is the required eq
approve if u like my ans
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