The focus of (x − h)² = 4p(y − k) is at: — (h, k + p).Focus is at (h, k + p). It is p units above the vertex (for upward parabola).
The directrix of (x − h)² = 4p(y − k) is: — y = k − p.Directrix: y = k − p. It is p units below the vertex (opposite direction from focus).
For x² = 8y, the focus is: — (0, 2).p = 2, vertex at origin. Focus at (h, k+p) = (0, 0+2) = (0, 2).
For x² = 8y, the directrix is: — y = −2.Directrix: y = k − p = 0 − 2 = −2.
For (y − 3)² = −12(x + 1), the parabola opens: — Left.The form (y−k)² = 4p(x−h) opens right if p > 0, left if p < 0. Here 4p = −12 so p = −3 < 0: opens left.
For a horizontal parabola (y−k)² = 4p(x−h), the focus is at: — (h + p, k).For a horizontal parabola, the focus is p units along the x-axis from the vertex: (h + p, k).
The latus rectum of a parabola has length: — 4p.The latus rectum (chord through focus, perpendicular to axis) has length |4p|.
The axis of symmetry of (x − 3)² = 12(y + 2) is: — x = 3.The axis of symmetry of this vertical parabola is the vertical line x = h = 3.
(y + 4)² = 8(x − 1): vertex is at: — (1, −4).Vertex of (y−k)² = 4p(x−h) is at (h, k) = (1, −4). (The form has y + 4, so k = −4.)
A parabolic dish has its vertex at origin, opens upward, and focus at (0, 3). Its equation is: — x² = 12y.p = 3, vertex at origin, vertical: x² = 4py = 4(3)y = 12y.
For a parabola with focus (3, 0) and directrix x = −3, the vertex is: — (0, 0).The vertex is midway between focus and directrix: midpoint of (3, 0) and (−3, 0) is (0, 0).
Write x² = −16y in standard form. The value of p is: — −4.4p = −16 → p = −4. Negative p means the parabola opens downward.
Focus of x² = −16y: — (0, −4).p = −4, vertex at origin. Focus at (0, k+p) = (0, 0+(−4)) = (0, −4).
Which equation represents a downward-opening parabola? — (x−1)² = −6(y+2).(x−1)² = −6(y+2): 4p = −6 so p < 0, which means the vertical parabola opens downward.
For x²/25 + y²/9 = 1, the foci are at: — (±4, 0).a² = 25, b² = 9. c² = 25 − 9 = 16, c = 4. Major axis is horizontal: foci at (±4, 0).
For x²/25 + y²/9 = 1, what is the sum of focal distances from any point on the ellipse? — 10.The sum of focal distances = 2a = 2(5) = 10.
For x²/16 + y²/25 = 1, the major axis is: — Vertical, length 10.a² = 25 is under y², so the major axis is vertical with length 2a = 10.
For x²/16 + y²/25 = 1, the foci are at: — (0, ±3).c² = 25 − 16 = 9, c = 3. Major axis is vertical: foci at (0, ±3).
Eccentricity e of an ellipse satisfies: — 0 < e < 1.Ellipse eccentricity e = c/a. Since c < a, we have 0 < e < 1. Closer to 0 = more circular.
A circle is a special ellipse where: — a = b.When a = b, the foci coincide at the center and the ellipse becomes a circle. Eccentricity = 0.
Vertices of x²/16 + y²/9 = 1 are at: — (±4, 0) and (0, ±3).Vertices are at the ends of the major axis (±a, 0) = (±4, 0) and co-vertices at (0, ±b) = (0, ±3).
(x−2)²/9 + (y+1)²/4 = 1: the semi-major axis length a = ? — 3.a² = 9, so a = 3. (9 > 4, so the larger denominator gives the major axis.)
If an ellipse has c = 3 and a = 5, then b = ? — 4.c² = a² − b² → 9 = 25 − b² → b² = 16 → b = 4.
Write the equation of an ellipse centered at (1, 2) with a = 5 (horizontal) and b = 3: — (x−1)²/25 + (y−2)²/9 = 1.Horizontal major axis: (x−h)²/a² + (y−k)²/b² = 1 = (x−1)²/25 + (y−2)²/9 = 1.
The eccentricity of a very elongated ellipse (c close to a) approaches: — 1.e = c/a. When c → a (very elongated), e → 1. A circle has e = 0.
What are the co-vertices of x²/25 + y²/16 = 1? — (0, ±4).Co-vertices are at the ends of the minor axis: (0, ±b) = (0, ±4).
An ellipse has a semi-major axis of 10 and eccentricity 0.6. Then c = ? — 6.e = c/a → 0.6 = c/10 → c = 6.
For x²/9 − y²/16 = 1, the foci are at: — (±5, 0).c² = a² + b² = 9 + 16 = 25, c = 5. Horizontal: foci at (±5, 0).
The asymptotes of x²/a² − y²/b² = 1 are: — y = ±(b/a)x.Asymptotes of horizontal hyperbola: y = ±(b/a)x.
For x²/9 − y²/16 = 1, the asymptotes are: — y = ±(4/3)x.a = 3, b = 4. Asymptotes: y = ±(b/a)x = ±(4/3)x.
How does the hyperbola c² = a² + b² differ from the ellipse c² = a² − b²? — For hyperbolas c > a; for ellipses c < a.For hyperbolas: c² = a² + b² means c > a (foci outside vertices). For ellipses: c² = a² − b² means c < a (foci inside).
The eccentricity e of a hyperbola satisfies: — e > 1.Hyperbola eccentricity e = c/a > 1 (since c > a). Compare: ellipse has e < 1, parabola has e = 1.
For y²/4 − x²/9 = 1, the vertices are at: — (0, ±2).Vertical hyperbola: vertices at (0, ±a) = (0, ±2).
Identify: 4x² − 9y² = 36 — Hyperbola.Divide by 36: x²/9 − y²/4 = 1. This is a hyperbola (subtraction between terms).
A hyperbola with a = 4, b = 3, horizontal orientation: c = ? — 5.c² = a² + b² = 16 + 9 = 25, c = 5.
(x−1)²/4 − (y+2)²/9 = 1: center is: — (1, −2).Center of (x−h)²/a² − (y−k)²/b² = 1 is at (h, k) = (1, −2).
The conjugate axis of x²/16 − y²/9 = 1 has length: — 6.The conjugate axis length = 2b = 2(3) = 6.
For x²/a² − y²/b² = 1, as |x| → ∞, the curve approaches: — y = ±(b/a)x.The asymptotes are y = ±(b/a)x for horizontal hyperbola.
Which is largest for a hyperbola: a, b, or c? — c.c² = a² + b² > a². So c > a always. c is always the largest.