(i) the focus is at (-6, 6) and the vertex is at (-2, 2)
Given:
Focus = (-6, 6)
Vertex = (-2, 2)
Let us find the slope of the axis (m1) = (6 - 2)/(-6 - (-2))
= 4/-4
= -1
Let us assume m2 be the slope of the directrix.
m1m2 = -1
-1m2 = -1
m2 = 1
Now, let us find the point on directrix.
(-2, 2) = [(x - 6/2), (y + 6)/2]
By equating we get,
(x - 6/2) = -2 and (y + 6)/2 = 2
x - 6 = -4 and y + 6 = 4
x = -4 + 6 and y = 4-6
x = 2 and y = -2
So the point of directrix is (2, -2).
We know that the equation of the lines passing through (x1, y1) and having slope m is y – y1 = m(x – x1)
y – (-2) = 1(x – 2)
y + 2 = x – 2
x – y – 4 = 0
Let us assume P(x, y) be any point on the parabola.
The distance between two points (x1, y1) and (x2, y2) is given as:
And the perpendicular distance from the point (x1, y1) to the line ax + by + c = 0 is
So by equating both, we get
Now by cross multiplying, we get
2x2 + 2y2 + 24x – 24y + 144 = x2 + y2 – 8x + 8y – 2xy + 16
x2 + y2 + 2xy + 32x – 32y + 128 = 0
∴ The equation of the parabola is x2 + y2 + 2xy + 32x – 32y + 128 = 0
(ii) the focus is at (0, -3) and the vertex is at (0, 0)
Given:
Focus = (0, -3)
Vertex = (0, 0)
Let us find the slope of the axis (m1) = (-3-0)/(0-0)
= -3/0
= ∞
Since the axis is parallel to the x-axis, the slope of the directrix is equal to the slope of x-axis = 0
So, m2 = 0
Now, let us find the point on directrix.
(0, 0) = [(x-0/2), (y-3)/2]
By equating we get,
(x/2) = 0 and (y-3)/2 = 0
x = 0 and y – 3 = 0
x = 0 and y = 3
So the point on directrix is (0, 3).
We know that the equation of the lines passing through (x1, y1) and having slope m is y – y1 = m(x – x1)
y – 3 = 0(x – 0)
y – 3 = 0
Now, let us assume P(x, y) be any point on the parabola.
The distance between two points (x1, y1) and (x2, y2) is given as:
And the perpendicular distance from the point (x1, y1) to the line ax + by + c = 0 is
So by equating both, we get
Now by cross multiplying, we get
x2 + y2 + 6y + 9 = y2 – 6y + 9
x2 + 12y = 0
∴ The equation of the parabola is x2 + 12y = 0
(iii) the focus is at (0, -3) and the vertex is at (-1, -3)
Given:
Focus = (0, -3)
Vertex = (-1, -3)
Let us find the slope of the axis (m1) = (-3-(-3))/(0-(-1))
= 0/1
= 0
We know, the products of the slopes of the perpendicular lines is – 1 for non – vertical lines.
Here the slope of the axis is equal to the slope of the x – axis. So, the slope of directrix is equal to the slope of y – axis i.e., ∞.
So, m2 = ∞
Now let us find the point on directrix.
(-1, -3) = [(x+0/2), (y-3)/2]
By equating we get,
(x/2) = -1 and (y-3)/2 = -3
x = -2 and y – 3 = -6
x = -2 and y = -6+3
x = -2 and y = -3
So, the point on directrix is (-2, -3)
We know that the equation of the lines passing through (x1, y1) and having slope m is y – y1 = m(x – x1)
y – (- 3) = ∞(x – (- 2))
(y+3)/ ∞ = x + 2
x + 2 = 0
Now, let us assume P(x, y) be any point on the parabola.
The distance between two points (x1, y1) and (x2, y2) is given as:
And the perpendicular distance from the point (x1, y1) to the line ax + by + c = 0 is
So by equating both, we get
By cross multiplying, we get
x2 + y2 + 6y + 9 = x2 + 4x + 4
y2 – 4x + 6y + 5 = 0
∴ The equation of the parabola is y2 – 4x + 6y + 5 = 0