Question -
Answer -
(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 thedirectrix.
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 thedirectrix 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 perpendicularlines 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
(iv) the focus is at (a, 0) and the vertex is at (a′, 0)
Given:
Focus = (a, 0)
Vertex = (a′, 0)
Let us find the slope of the axis (m1) =(0-0)/(a′, a)
= 0/(a′, a)
= 0
We know, the products of the slopes of the perpendicularlines 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.
(a′, 0)= [(x+a/2), (y+0)/2]
By equating we get,
(x+a/2) = a′ and (y)/2 = 0
x + a = 2a′ and y = 0
x = (2a′ – a) and y = 0
So the point on directrix is (2a′ – a, 0).
We know that the equation of the lines passing through (x1,y1) and having slope m is y – y1 = m(x – x1)
y – (0) = ∞(x – (2a′ – a))
y/∞ = x + a – 2a′
x + a – 2a′ = 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 – 2ax + a2 =x2 + a2 + 4(a′)2 + 2ax – 4aa′ –4a′x
y2 – (4a – 4a′)x + a2 –4(a′)2 + 4aa′ = 0
∴ The equation of the parabola is y2 – (4a –4a′)x + a2 – 4(a′)2 + 4aa′ = 0
(v) the focus is at (0, 0) and the vertex is at the intersectionof the lines x + y = 1 and x – y = 3.
Given:
Focus = (0, 0)
Vertex = intersection of the lines x + y = 1 and x – y = 3.
So the intersecting point of above lines is (2, -1)
Vertex = (2, -1)
Slope of axis (m1) = (-1-0)/(2-0)
= -1/2
We know that the products of the slopes of the perpendicularlines is – 1.
Let us assume m2 be the slope of thedirectrix.
m1.m2 = -1
-1/2 . m2 = -1
So m2 = 2
Now let us find the point on directrix.
(2, -1) = [(x+0)/2, (y+0)/2]
(x)/2 = 2 and y/2 = -1
x = 4 and y = -2
The point on directrix is (4, – 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) = 2(x – 4)
y + 2 = 2x – 8
2x – y – 10 = 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
5x2 + 5y2 = 4x2 +y2 – 40x + 20y – 4xy + 100
x2 + 4y2 + 4xy + 40x – 20y –100 = 0
∴ The equation of the parabola is x2 + 4y2 +4xy + 40x – 20y – 100 = 0