Exercise 6.2
In Exercise 1 - 8, (a) sketch the parabola, (b) show the focus, vertex, directrix and end points of latus rectum.
1. y2 = 8x
Solution
y2 = 8x
Comparing with y2 = 4px,
4px = 8x
p = 8x ⁄ 4x
p = 2
Therefore, the vertex V = (0,0),
the focus F = (p, 0) = (2, 0),
the directrix is x = -p = -2,
end points of latus rectum are
(p, -2p), (p, 2p) = (2, -4), (2, 4)
2. y2 = -6x
Solution
y2 = -6x
Comparing with y2 = -4px,
-4px = -6x
p = -6x ⁄ -4x
= p = 3 ⁄ 2 = 1.5
Therefore, the vertex V = (0,0),
the focus F = (-p, 0) = (-1.5, 0),
the directrix is x = p = 1.5,
end points of latus rectum are
(-p, -2p), (-p, 2p) = (-1.5, 3), (-1.5, -3)
3. x + y2 = 0
Solution
x + y2 = 0
y2 = -x
Comparing with y2 = -4px,
-4px = -x
p = -x ⁄ -4x
p = 1 ⁄ 4 = 0.25
Therefore, the vertex V = (0,0),
the focus F = (-p, 0) = (-0.25, 0),
the directrix is x = p = 0.25,
end points of latus rectum are
(-p, -2p), (-p, 2p) = (-0.25, -0.5), (-0.25, 0.5)
4. y = x2
Solution
y = x2
x2 = y
Comparing with x2 = 4py,
4py = y
p = y ⁄ 4y
= -1 ⁄ 4
Therefore, the vertex V = (0,0),
the focus F = (0, p) = (0, 1 ⁄ 4),
the directrix is y = -p = - 1 ⁄ 4,
end points of latus rectum are
(-2p, p), (2p, p) = (- 1 ⁄ 2,
1 ⁄ 4), (1 ⁄ 2,
1 ⁄ 4)
5. y = -x2
Solution
y = -x2
x2 = -y
Comparing with x2 = -4py,
-4py = -y
p = -y ⁄ -4y
= 1 ⁄ 4
Therefore, the vertex V = (0,0),
the focus F = (0, -p) = (0, - 1 ⁄ 4),
the directrix is y = p = - 1 ⁄ 4,
end points of latus rectum are
(-2p, -p), (2p, -p) = (- 1 ⁄ 2, -
1 ⁄ 4), (1 ⁄ 2, -
1 ⁄ 4)
6. y = 3x2
Solution
y = 3x2
x2 = y ⁄ 3
Comparing with x2 = 4py,
4py = y ⁄ 3
p = y ⁄ 3⋅4y
= 1 ⁄ 12
Therefore, the vertex V = (0,0),
the focus F = (0, p) = (0, 1 ⁄ 12),
the directrix is y = -p = - 1 ⁄ 12,
end points of latus rectum are
(-2p, p), (2p, p) = (- 1 ⁄ 6,
1 ⁄ 12), (1 ⁄ 6,
1 ⁄ 12)
7. y = -3x2
Solution
y = -3x2
x2 = - y ⁄ 3
Comparing with x2 = -4py,
-4py = - y ⁄ 3
p = -y ⁄ 3⋅(-4)y
= 1 ⁄ 12
Therefore, the vertex V = (0,0),
the focus F = (0, -p) = (0, - 1 ⁄ 12),
the directrix is y = p = 1 ⁄ 12,
end points of latus rectum are
(-2p, -p), (2p, -p) = (- 1 ⁄ 6, -
1 ⁄ 12), (1 ⁄ 6, -
1 ⁄ 12)
8. x2 + 6y = 0
Solution
x2 = -6y
Comparing with x2 = -4py,
-4py = - 6y
p = -6y ⁄ -4y
= 3 ⁄ 2
Therefore, the vertex V = (0,0),
the focus F = (0, -p) = (0, - 3 ⁄ 2),
the directrix is y = p = 3 ⁄ 2,
end points of latus rectum are
(-2p, -p), (2p, -p) = (-3 , -
3 ⁄ 2), (3, -
3 ⁄ 2)
In Exercise 9 - 16, find an equation for the parabola satisfying the given conditions.
9. Vertex (0, 0), focus (3, 0).
Solution
Vertex (0,0),
focus (3, 0) = (p, 0)
p = 3
The equation of the parabola is
y2 = 4px
= 4(3)x
= 12x
10. Vertex (0, 0), focus (0, -4)
Solution
Vertex (0, 0)
focus (0, -4) = (0, -p)
-p = -4
p = 4
The equation of the parabola is
x2 = -4py
= -4 (4)y
= -16y
11. Vertex (0, 0). directrix x = 7.
Solution
Vertex (0, 0)
directrix x = 7
∴ p = 7
The equation of the parabola is
y2 = -4px
= -4 (7)x
= -28x
12. Vertex (0, 0), directrix y = 1 ⁄ 2.
Solution
Vertex (0, 0)
directrix y = 1 ⁄ 2.
∴ p = 1 ⁄ 2.
The equation of the parabola is
x2 = -4py
= -4 (1 ⁄ 2)y
= -2y
13. Focus (0, -3), directrix y = 3.
Solution
Focus (0, -3),
directrix y = 3
∴ p = 3
The equation of the parabola is
x2 = -4py
= (-4)⋅3 y
= -12y
14. Focus (6, 0), directrix x = -6
Solution
Focus (6, 0) = (p, 0)
∴ p = 6
directrix x = -6
x = -p
-p = -6
p = 6
The equation of the parabola is
y2 = 4px
= 4⋅6x
= 24x
15. Vertex (0, 0); symmetric about the x-axis; passes through (2, 2).
Solution
Since the parabola is symmetric about the x-axis, Vertex V = (0, 0) and passes through (2, 2), the equation of the form is
y2 = 4px
At (2, 2)
22 = 4p(2)
4 ⁄ 2 = 4p
4p = 2
Threrfore, the equation of the parabola is
y2 = 4px
= 2x (∵ 4p = 2)
16. Vertex (0, 0); symmetric about the y-axis; passes through (-1, 3).
Solution
Vertex V = (0, 0)
x2 = 4py
At (-1, 3)
(-1)2 = 4p(3)
1 ⁄ 3 = 4p
Threrfore, the equation of the parabola is
x2 = 4py
x2 = 1 ⁄ 3y (∵ 4p =
1 ⁄ 3)
3x2 = y
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