For Exercises 51-52, a. Determine the eccentricity of each hyperbola. b. Based on the eccentricity, match the equation with its graph. The scaling is the same for both graphs. Equation 1: x 2 144 − y 2 81 = 1 Equation 2: x 2 81 − y 2 144 = 1
For Exercises 51-52, a. Determine the eccentricity of each hyperbola. b. Based on the eccentricity, match the equation with its graph. The scaling is the same for both graphs. Equation 1: x 2 144 − y 2 81 = 1 Equation 2: x 2 81 − y 2 144 = 1
3. Reason Jupiter's gravity changes a spacecraft's
path to a hyperbola as it approaches. The
focus of the hyperbola nearest the spacecraft's
path is the center of Jupiter. The diameter of
the planet is 139,822 km. Suppose the path
of an approaching spacecraft has an equation
with a = 80,000 km and c= 170,000 Km.
Assuming the transverse axis.js horizontal,
write the equation that models the path of the
spacecraft as it approaches Jupiter. What is the
distance from the spacecraft to the planet at
the vertex of the hyperbola?
I.
Direction: Write the following equations in standard form and find the (a) vertex, (b)
focus, (c) length of latus rectum, and (d) equation of the directrix.
1. x²-2x-24y+1=0
Standard Form Equation:
a. Vertex:
b. Focus:
c. Length of LR:
d. Directrix:
3. x²-6x-y+3=0
Standard Form Equation:
a. Vertex:
b. Focus:
c. Length of LR:
d. Directrix:
PCLNI1S
5. y² +2x-4y+4=0
Standard Form Equation:
a. Vertex:
b. Focus:
c. Length of LR:
d. Directrix:
Written Works
2. x² +8x+48y+64=0
Standard Form Equation:
a. Vertex:
b. Focus:
c. Length of LR:
d. Directrix:
4. y²-8x+10y+57=0
Standard Form Equation:
a. Vertex:
b.
Focus:
c. Length of LR:
d. Directrix:
2nd Quarter
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Finding The Focus and Directrix of a Parabola - Conic Sections; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=KYgmOTLbuqE;License: Standard YouTube License, CC-BY