Conic Sections, Translation of Axes

Translation of Axes

If a parabola with vertex at the origin and axis of symmetry along a coordinate axis is shifted horizontally h units and then vertically k units, then we get the parabola with vertex at (h, k) and axis of symmetry parallel to a coordinate axis.

In the figure, a new x'y'-coordinate system appears with the origin O' at the point (h, k).
A point P in the plane has both (x, y) coordinates and (x', y') coordinates.

In the figure, their coordinates are related by x' = x - h, y' = y - k or equivalently,
x = x' + h, y = y' + k.
These are called Translation equations
As an example, if a new origin is (h, k) and the xy-coordinates of the point P are (2, 5), then the x'y'-coordinates of P are
x' = x - h = 2 - 4 = -2 and
y' = y - k = 5 - (-1) = 6 .
Therefore the new coordinates of the point is (-2, 6).

Translated Parabola
Now we can consider a parabola with vertex V at (h, k) in any xy-coordinate system and axis of symmetry parallel to the y-axis. If we translate the axis, the vertex V will be at the origin of an x'y'-coordinate system.

Parabola with Vertex (h, k) and Axis parallel to y-axis
In x'y'-coordinates, the equation of the parabola will be
(x')² = 4py' or (x')² = -4py' (See the figure).
fig 6.3-3

The corresponding equation in xy-coordinates will be
(x - h)² = ±4p(y - k).

Here the (+) sign occurs if the parabola opens in the positive y direction and the (-) sign occurs if the parabola opens in the negative y directioin.

Parabola with Vertex (h, k) and Axis parallel to x-axis
The equation of the parabola with vertex (h, k) and axis parallel to x-axis is of the form
(y - k)² = ±4p(x - h).

Here the (+) sign occurs if the parabola opens in the positive x direction and the (-) sign occurs if the parabola opens in the negative x direction.
The four possible such orientations whose vertices not at the origin are shown in the following figures. These are called the standard positions of a parabola, and the resulting equations are called the standard forms of equation of a parabola.

Equations of a Parabola:
Vertex at (h, k), Parallel to Coordinate Axis, p>0

Equation	Vertex	Focus	Directrix	Axis of Symmetry	Description
(y - k)² = 4p(x - h)	(h, k)	(h+p, k)	x = h - p	horizontal axis, y = k	opens to the right
(y - k)² = -4p(x - h)	(h, k)	(h-p, k)	x = h + p	horizontal axis, y = k	opens to the left
(x - h)² = 4p(y - k)	(h, k)	(h, k+p)	y = k - p	vertical axis, x = h	opens up
(x - h)² = -4p(y - k)	(h, k)	(h, k-p)	y = k + p	vertical axis, x = h	opens down

Example 8.
Write the given equation y² - 12x - 6y - 3 = 0 in standard form. Sketch its graph showing the vertex, focus, directrix and end points of latus rectum.
Solution
y² - 12x - 6y - 3 = 0
y² - 6y = 12x + 3
y² - 6y + 3² = 12x + 3 + 9
(y - 3)² = 12x + 12
(y - 3)² = 12(x + 1)
Comparing with (y - k)² = 4p(x - h), we have h = -1, k = 3 and p = 3.
Therefore, the vertex V = (-1, 3), the focus F = (2, 3) and the directrix is x = -4.

Example 9.
Write the given parabola equation x² - 2x + 8y - 23 = 0 in standard form. Sketch its graph showing the vertex, focus, directrix and endpoints of latus rectum.
Solution
x² - 2x + 8y - 23 = 0
x² - 2x = -8y + 23
(x - 1)² = -8y + 24
(x - 1)² = -8(y - 3)
Comparing with (x - h)² = -4p(y - k), we have h = 1, k = 3 and p = 2.
Therefore, the vertex V = (1, 3), the focus F = (1, 1) and the directrix is y = 5.

Example 10.
Find the standard form of the equation of the parabola with vertex (1, 2) and focus at (4, 2).
Solution
Since the focus and vertex are on a horizontal line and the focus is on the right of the vertex, the parabola opens to the right.
The vertex V = (1, 2) , the focus F = (4, 2) = (1+3, 2).
Therefore h = 1, k = 2 and p = 3.
The equation of the parabola is
(y - k)² = 4p(x - h)
(y - 2)² = 4 × 3(x - 1)
(y - 2)² = 12(x - 1)

Example 11.
Find the general form of the equation of the parabola with vertex (2, 1) and focus (2, 4).
Solution
Since the focus and vertex are on a vertical line and the focus is above the vertex, the parabola opens up.
Then V = (h, k) = (2, 1), F = (2, 4) = (2, 1+3) and p = 3.
The equation of the parabola is
(x - h)² = 4p(y - k)
(x - 2)² = 4 × 3(y - 1)
x² - 4x + 16 = 12y
x² - 4x - 12y + 16 = 0.
The general form of the parabola is x² - 4x - 12y + 16 = 0.

Exercise 6.3

In Exercise 1 - 6, (a) sketch the parabola, (b) show the focus, vertex, directrix and end points of latus rectum.
1. x² + 4x + 4y = 0
Solution
x² + 4x + 4y = 0
x² + 4x = -4y
x² + 4x + 4 - 4 = -4y
x² + 4x + 4 = -4y + 4
(x + 2)² = -4(y - 1)
Comparing with (x - h)² = -4p(y - k),
-h = 2, -4p = -4, -k = -1
h = -2, p = 1, k = 1
Therefore, the vertex V = (h, k) = (-2, 1),
the focus F = (h, k-p) = (-2, 0),
the directrix is y = k + p = 1 + 1 = 2,
end points of latus rectum are
(h-2p, k-p), (h+2p, k-p) = (-4,0), (0, 0)

2. (x - 1)² + 8(y + 1) = 0
Solution
(x - 1)² + 8(y + 1) = 0
(x - 1)² = -8(y + 1)
Comparing with (x-h)² = -4p(y-k),
-h = -1, -4p = -8, -k = 1
h = 1, p = 2, k = -1
Therefore, the vertex V = (h, k) = (1, -1),
the focus F = (h, k-p) = (1, -3),
the directrix is y = k + p = -1 + 2 = 1
end points of latus rectum are (h-2p, k-p), (h+2p, k-p) = (-3, -3), (5, -3)

3. (x + 5) + (y - 1)² = 0
Solution
(x + 5) + (y - 1)² = 0
(y - 1)² = -(x + 5)
Comparing with (y - k)² = -4p(x - h),
-k = -1, -4p = -1, -h = 5
k = 1, p= ¹ ⁄ ₄, h = -5
Therefore, the vertex V = (h, k) = (-5, 1),
the focus F = (h-p, k) = (-5- ¹ ⁄ ₄, 1) = (^-21 ⁄ ₄, 1)
= (-5.25, 1)
Directrix is x = h + p = -4.75
end points of latus rectum are
(h-p, k-2p), (h-p, k+2p) = (^-21 ⁄ ₄, ¹ ⁄ ₂), (^-21 ⁄ ₄, ³ ⁄ ₂)
= (-5.25, 0.5), (-5.25, 1.5)

4. y² + 6y + 8x + 25 = 0
Solution
y² + 6y + 8x + 25 = 0
y² + 6y = -8x - 25
y² + 6y + 9 - 9 = -8x - 25
y² + 6y + 9 = -8x - 25 + 9
(y + 3)² = -8x - 16
(y + 3)² = -8(x + 2)
Comparing with (y - k)² = -4p(x - h),
-k = 3, -4p = -8, -h = 2
k = -3, p = 2, h = -2
Therefore, the vertex V = (h, k) = (-2,-3),
the focus F = (h-p, k) = (-4, -3),
the directrix is x = h + p = -2 + 2 = 0 ,
end points of latus rectum are
(h-p, k-2p), (h-p, k+2p) = (-4, -7), (-4, 1)

5. x² + 4x + 6y - 2 = 0
Solution
x² + 4x + 6y - 2 = 0
x² + 4x = -6y + 2
x² + 4x + 4 - 4 = -6y + 2
x² + 4x + 4 = -6y + 2 + 4
(x+2)² = -6y + 6
(x + 2)² = -6(y - 1)
Comparing with (x - h)² = -4p(y - k),
-h = 2, -4p = -6, -k = -1
h = -2, p = ^-6 ⁄ _-4 = ³ ⁄ ₂, k = 1
Therefore, the vertex V = (h, k) = (-2, 1),
the focus F = (h, k-p) = (-2, 1 - ³ ⁄ ₂) = (-2, ^-1⁄₂) = (-2, -0.5)
the directrix is y = k + p = 1 + ³ ⁄ ₂ = ⁵⁄₂ = 2.5,
end points of latus rectum are
(h-2p, k-p), (h+2p, k-p) = (-2 - 2(³ ⁄ ₂), 1 - ³ ⁄ ₂), (-2 + 2(³ ⁄ ₂), 1 -³ ⁄ ₂) = (-5, ^-1⁄₂), (1, ^-1⁄₂)
= (-5, -0.5), (1, -0.5)

6. y² + x + y = 0
Solution
y² + x + y = 0
y² + y = -x
y² + y + ¹ ⁄ ₄ - ¹⁄4 = -x
y² + y + ¹ ⁄ ₄ = -x + ¹⁄4
(y + ¹ ⁄ ₂)² = -1 (x - ¹⁄4)
Comparing with (y - k)² = -4p(x - h),
-k = ¹ ⁄ ₂, -4p = -1, -h = ^-1 ⁄ ₄
k = ^-1 ⁄ ₂ , p = ¹ ⁄ ₄ , h = ¹⁄4
Therefore, the vertex V = (h, k) = (¹ ⁄ ₄, ^-1⁄2) = (0.25, -0.5) ,
the focus F = (h-p, k) = (0, ^-1 ⁄ ₂) = (0, -0.5),
the directrix is x = h + p = ¹ ⁄ ₄ + ¹ ⁄ ₄ = ²⁄4 = ¹⁄₂ = 0.5
end points of latus rectum are
(h-p, k+2p), (h-p, k-2p) = (0, 0), (0, -1)

In Exercise 7 - 14, find an equation for the parabola satisfying the given conditions.
7. Vertex (4, -5),
Focus (1, -5)
Solution
Since the focus and vertex are on a horizontal line and the focus is to the left of the vertex, the parabola is opens to the left.
The vertex V = (h, k) = (4, -5)
h = 4, k = -5
The focus F = (h-p, k) = (1, -5)
h - p = 1
4 - p = 1
-p = -3
p = 3
The equation of the parabola is
(y - k)² = -4p(x -h)
(y - (-5))² = -4(3) (x - 4)
(y + 5)² = -12(x - 4)

8. Vertex (5, 1)
focus (5, 3)
Solution
Since the focus and vertex are on a vertical line and the focus is above the vertex, the parabola opens up.
The vertex V = (h, k) = (5, 1)
h = 5, k = 1
the focus F = (h, k+p) = (5, 3)
k + p = 3
1 + p = 3
p = 2
The equation of the parabola is
(x - h)² = 4p(y - k)
(x - 5)² = 4(2)(y - 1)
= 8(y - 1)

9. Vertex (1, 1),
directrix y = -2.
Solution
Since the vertex is (1, 1) and directrix is parallel to the x-axis, the parabola is opens up.
the vertex V = (h, k) = (1, 1)
h = 1,
k = 1
directrix y = -2
y = k - p
k - p = -2
1 - p = -2
p = 3
focus F = (h, k+p) = (1, 1+3) = (1, 4)
The equation of the parabola is
(x - h)² = 4p(y - k)
(x - 1)² = 12(y - 1)

10. Vertex (2, -1). directrix x = -1.
Solution
Since the vertex is (2, -1) and the directrix is parallel to the y-axis, the parabola opens to the right.
vertex V = (h, k) = (2, -1)
h = 2, k = -1
directrix x = -1
x = h - p
h - p = -1
2 - p = -1
-p = -3
p = 3
The equation of the parabola is
(y - k)² = 4p(x - h)
(y - (-1))² = 4 (3) (x - 2)
(y + 1)² = 12(x - 2)

11. Focus (-1, 4), directrix x = 5.
Solution
Since the focus is (-1, 4) and the directrix x = 5 is parallel to the y-axis, the parabola opens to the left.
focus F = (h-p, k) = (-1, 4)
h - p = -1
h = p - 1
k = 4
directrix: x = 5
directrix x = h + p
h + p = 5
p - 1 + p = 5
2p = 6
p = 3

h + p = 5
h + 3 = 5
h = 2
The equation of the parabola is
(y - k)² = -4p(x - h)
(y - 4)² = -4(3) (x -2)
(y - 4)² = -12(x - 2)

12. Focus (6, 0), directrix x = -6.
Solution
Since the focus is (6, 0) and the directrix x = -6 is parallel to the y-axis, the parabola opens to the right.
focus F(6, 0) = (h+p, k)
h+p = 6, k = 0
h = 6 - p
directrix x = -6
directrix x = h - p
h - p = -6
6 - h - p= -6
-2p = -12
p = 6

h = 6 - p
h = 6 - 6
h = 0
The equation of the parabola is
(y - k)² = 4p(x - h)
(y - 0)² = 4(6)(x - 0)
y² = 24x

13.Axis x = 0 passes through (2, -1) and (-4, 5).
Solution
Axis x = 0 is symmetric line.
Axis x = 0 passes through (2, -1) and (-4, 5)
The parabola is open up.
the vertex V = (h, k) = (0, k)
h = 0
The equation of the parabola is
(x - h)² = 4p(y - k)
(x - 0)² = 4p(y - k)
x² = 4p(y - k)
At (2, -1)
x² = 4p(y - k)
2² = 4p(-1 - k)
-4p - 4pk = 4 ______(1)
At (-4, 5),
x² = 4p(y - k)
(-4)² = 4p(5 - k)
16 = 20p - 4pk
20p - 4pk = 16 ______(2)
Equation (2) - (1),

p = ¹²⁄₂₄ = ¹⁄₂
Substituting p = ¹⁄₂ in equation (1)
-4p - 4pk = 4
-4 ⋅ ¹⁄₂ - 4 ⋅ ¹⁄₂k = 4
-2 - 2k = 4
-2k = 6
k = -3
∴ The equation of the parabola is
x² = 4p(y - k)
x² = 4(¹⁄₂)(y - (-3))
x² = 2(y + 3)

14. Axis y = 0 passes through (3, 2) and (2, -3).
Solution
Axis y = 0 is symmetric line.
Axis y = 0 passes through (3, 2) and (2, -3).
The parabola is open left.
the vertex V = (h, k) = (h, 0)
k = 0
The equation of the parabola is
(y - k)² = -4p(x - h)
(y - 0)² = -4p(x - h)
y² = -4p(x - h)
At (3, 2),
2² = -4p(3 - h)
-12p + 4pk = 4 ______(1)
At (2, -3),
y² = -4p(x - h)
(-3)² = -4p(2 - h)
9 = -8p + 4ph
-8p + 4ph = 9 ______(2)
Equation (2) - (1),

p = ⁵ ⁄ ₄
Substituting p = ⁵ ⁄ ₄ in equation (1)
-12p + 4ph = 4
-12(⁵ ⁄ ₄) + 4(⁵⁄₄)h = 4
-15 + 5h = 4
5h = 19
h = ¹⁹⁄₅
∴ The equation of the parabola is
y² = -4p(x - h)
= -4(⁵⁄₄)(x - ¹⁹⁄₅)
= -5(x - ¹⁹⁄₅)

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