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find the equation of parabola , when the vertex is(2,1) and directrix is x=y-1
Dear student Equation of directrix= x - y+1=0--------(i) Let the coordinate of focus = (a,b) the axis of parabola is perpendicular to directrix. So the equation of axis of parabola may be taken as x+y+k=0 it passes through (2,1)⇒ 2+1+k=0⇒ k=-3 ⇒ the equation of axis of parabola is x+y-3=0---------------(ii) Now, for the point of intersection of directrix and axis of parabolaOn solving (i) and (ii), we get, y = 2put the value of y in (i), we get, x = 1 Thus, the point of intersection of directrix and axis of parabola (X, Y) = (1, 2)Coordinates of z = (1,2) Vertex A (2,1) is the mid-point of focus and z = (1,2)2 = (a+1)/2a = 3 1 = (b+2)/2b = 0 Coordinate of focus = (3,0) Now you can simply get the equation of parabola byPS = PM
Equation of directrix= x - y+1=0--------(i)
Let the coordinate of focus = (a,b)
the axis of parabola is perpendicular to directrix.
So the equation of axis of parabola may be taken as x+y+k=0
it passes through (2,1)
⇒ 2+1+k=0
⇒ k=-3
⇒ the equation of axis of parabola is x+y-3=0---------------(ii)
Now, for the point of intersection of directrix and axis of parabola
On solving (i) and (ii), we get,
y = 2
put the value of y in (i), we get,
x = 1
Thus, the point of intersection of directrix and axis of parabola
(X, Y) = (1, 2)
Coordinates of
z = (1,2)
Vertex A (2,1) is the mid-point of focus and
2 = (a+1)/2
a = 3
1 = (b+2)/2
b = 0
Coordinate of focus = (3,0)
Now you can simply get the equation of parabola by
PS = PM
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