Math: Conic Sections: Hyperbolas and Parabolas
Graphing, standard forms, foci, directrices, and asymptotes
Math: Conic Sections: Hyperbolas and Parabolas
Graphing, standard forms, foci, directrices, and asymptotes
Math - Grade 9-12
- 1
For the parabola (x - 2)^2 = 8(y + 1), find the vertex, focus, directrix, and direction of opening.
Compare the equation to (x - h)^2 = 4p(y - k).
The vertex is (2, -1). Since 4p = 8, p = 2, so the focus is (2, 1), the directrix is y = -3, and the parabola opens upward. - 2
Write the equation of the parabola with vertex (-3, 4) and focus (-3, 1).
A vertical parabola has the form (x - h)^2 = 4p(y - k).
The focus is 3 units below the vertex, so p = -3. The equation is (x + 3)^2 = -12(y - 4). - 3
Rewrite y = x^2 - 6x + 11 in vertex form. Then state the vertex and direction of opening.
The equation in vertex form is y = (x - 3)^2 + 2. The vertex is (3, 2), and the parabola opens upward because the coefficient of the squared term is positive. - 4
Write the equation of the parabola with focus (5, 2) and directrix x = 1.
The vertex is halfway between the focus and directrix at (3, 2). Since p = 2, the equation is (y - 2)^2 = 8(x - 3). - 5
For the equation y^2 + 4y = 12x - 8, put the parabola in standard form and find the vertex, focus, and directrix.
Complete the square on the y terms before comparing to (y - k)^2 = 4p(x - h).
Completing the square gives (y + 2)^2 = 12(x - 1/3). The vertex is (1/3, -2), p = 3, the focus is (10/3, -2), and the directrix is x = -8/3. - 6
For the hyperbola (x - 1)^2/9 - (y + 2)^2/16 = 1, find the center, vertices, foci, and equations of the asymptotes.
The center is (1, -2). Since a = 3, b = 4, and c = 5, the vertices are (-2, -2) and (4, -2), the foci are (-4, -2) and (6, -2), and the asymptotes are y + 2 = (4/3)(x - 1) and y + 2 = -(4/3)(x - 1). - 7
Write the equation of the hyperbola with center (0, 0), vertices (0, 5) and (0, -5), and foci (0, 13) and (0, -13).
For a hyperbola, c^2 = a^2 + b^2.
The hyperbola opens up and down, so its form is y^2/a^2 - x^2/b^2 = 1. Since a = 5 and c = 13, b^2 = 13^2 - 5^2 = 144, so the equation is y^2/25 - x^2/144 = 1. - 8
Find the equations of the asymptotes for the hyperbola (y - 3)^2/4 - (x + 1)^2/25 = 1.
The center is (-1, 3), a = 2, and b = 5. Because the hyperbola is vertical, the asymptotes are y - 3 = (2/5)(x + 1) and y - 3 = -(2/5)(x + 1). - 9
Put 9x^2 - 16y^2 - 54x - 64y - 127 = 0 in standard form. Then state the center and transverse axis direction.
Group the x terms and y terms separately, then complete the square for each group.
Completing the square gives 9(x - 3)^2 - 16(y + 2)^2 = 144, so the standard form is (x - 3)^2/16 - (y + 2)^2/9 = 1. The center is (3, -2), and the transverse axis is horizontal. - 10
Classify x^2 - 4x - 8y + 20 = 0 as a parabola or hyperbola. Then write it in standard form and state its vertex.
This equation represents a parabola because only one variable is squared. Completing the square gives (x - 2)^2 = 8(y - 2), so the vertex is (2, 2). - 11
A hyperbola has foci (-4, 0) and (4, 0). For every point on the hyperbola, the absolute difference of the distances to the foci is 6. Write the equation in standard form.
For a hyperbola, the distance difference equals 2a.
The center is (0, 0), c = 4, and 2a = 6, so a = 3. Since b^2 = c^2 - a^2 = 16 - 9 = 7, the equation is x^2/9 - y^2/7 = 1. - 12
A parabolic reflector has cross-section y = (1/12)x^2 with vertex at the origin. Find the focus of the parabola.
Compare x^2 = 12y to x^2 = 4py.
Rewrite the equation as x^2 = 12y. Since 4p = 12, p = 3, so the focus is (0, 3).