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Find the equation of the parabola, if
(i) the focus is at (-6, 6) and the vertex is at (-2, 2)

(ii) the focus is at (0, -3) and the vertex is at (0, 0)

(iii) the focus is at (0, -3) and the vertex is at (-1, -3)

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(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

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