Chapter 10 Conics 10.2 Circles Standard Form of a Circle General

Chapter 10 Conics
10.2 Circles
Standard Form of a Circle
General Form
Graph
Ex 1: Write the standard form of the equation of the circle that has
a. Center at (2,4) and passes through (βˆ’8, 15)
b. Center at (1, βˆ’2)and is tangent to the line 𝑦 = 6
Ex 2: Rewrite π‘₯ 2 + 𝑦 2 βˆ’ 4π‘₯ + 14𝑦 βˆ’ 47 = 0 into standard form.
Ex 3: Write the equation of the circle in standard form that passes through (0, βˆ’1), (βˆ’3, βˆ’2), (βˆ’6, βˆ’1)
10.3 Ellipse
Standard Form of Ellipse
Ex 1: Graph
(π‘₯βˆ’1)2
16
+
(𝑦+2)2
36
=1
Ex 2: Graph
foci.
(π‘₯βˆ’1)2
25
+
(π‘¦βˆ’1)2
4
= 1 and state the coordinates of the
Eccentricity of an Ellipse
Ex 3: Write 16π‘₯ 2 + 25𝑦 2 βˆ’ 96π‘₯ βˆ’ 200𝑦 + 144 = 0 in standard form.
Ex 4: Write the equation of an ellipse in standard form given it has
a. Center (2,5), vertical ellipse, semi minor axis 4 units, major axis is 20 units
b. Center at (βˆ’4,2), foci at (βˆ’4, 2 ± √13), minor vertex at (βˆ’5,2) and (βˆ’3,2)
1
c. Major vertices at (7,15) and (7,3), eccentricity = 4
10.4 Hyperbola
Standard Form of Hyperbolas
Ex 1: Graph
(π‘¦βˆ’2)2
4
+
(π‘₯βˆ’1)2
16
=1
Ex 2: Write βˆ’4π‘₯ 2 + 9𝑦 2 βˆ’ 24π‘₯ βˆ’ 90𝑦 + 153 = 0 in standard form.
Ex 3: Write the equation of a hyperbola in standard form given that it has
a. Vertices at (2, βˆ’1) and (βˆ’8, βˆ’1), conjugate axis is 4 units long
b. Foci (2, βˆ’3) and (2,7), vertices (2, βˆ’1) and (2,5)
10.6 Parabolas
Standard form of Parabolas
Ex 1: Graph 3π‘₯ 2 βˆ’ 30𝑦 βˆ’ 18π‘₯ + 87 = 0. Identify the
coordinates of the vertex, focus, the equation of the directrix and
the axis of symmetry.
Ex 2: Graph (π‘¦βˆ’1)2 = 8(π‘₯ + 3). Identify vertex, focus, directrix
and axis of symmetry.
Ex 3: Write the equation of the parabola in standard form given that it’s focus is at (4, βˆ’2) and the equation
of the directrix is 𝑦 = 6
10.6 Rectangular and Parametric Forms of Conic Sections
Identifying Conic Sections from general form
Ex 1: Identify the conic section represented by each equation
a. 7π‘₯ 2 + 2𝑦 2 βˆ’ 4π‘₯ + 51𝑦 βˆ’ 65 = 0
b. βˆ’10π‘₯ 2 + 8𝑦 2 βˆ’
25
3
1
π‘₯ + 2 𝑦 βˆ’ √7 = 0
c. βˆ’5π‘₯ 2 βˆ’ 5𝑦 2 βˆ’ 16π‘₯ + 4 = 0
d. π‘₯ 2 + 𝑦 βˆ’ 3π‘₯ + 4 = 0