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When deriving the equation of a hyperbola using foci and the distance formula, the term involving in the equation is typically squared because:
It makes the equation symmetric
It simplifies the calculations
The hyperbola can be oriented either horizontally or vertically
Given a hyperbola with foci at (0,2) and (0,−2) and a distance between the foci of 4, the equation of the hyperbola is:
x2−16y2=1x^2-16y^2=1x2−16y2=1
x2+16y2=1x^2+16y^2=1x2+16y2=1
For a hyperbola with foci at (0,3) and (0,−3), and a distance between the foci of 6, the equation of the hyperbola is:
9x2−16y2=139x^2-16y^2=139x2−16y2=13
6x2−25y2=16x^2-25y^2=16x2−25y2=1
9x2−16y2=09x^2-16y^2=09x2−16y2=0
The distance between the foci of a hyperbola is 2c. The distance formula to derive the equation equal to:
2a
2b
2c
In the distance formula, the variables x1 and y1 represent:
The coordinates of the center of the hyperbola
the coordinates of one focus of the hyperbola
The coordinates of the other focus of the hyperbola
When using the distance formula to find the distance between a point on a hyperbola and one of its foci, the result should be equal to:
The length of the major axis
the length of the minor axis
The distance between the foci
The distance formula is used to calculate the distance between:
The two foci of a hyperbola
the center and a vertex of a hyperbola
A point on a hyperbola and one of its foci
The distance formula can be used to find the distance between:
The center and a vertex of a hyperbola
the center and a focus of a hyperbola
Two points on the minor axis of a hyperbola
If the distance between the foci of a hyperbola is 10 and a point on the hyperbola is (5,3), the equation for the distance using the distance formula is:
(x−5)2+(y−3)2=10\left(x-5\right)^2+\left(y-3\right)^2=10(x−5)2+(y−3)2=10
(x−5)2−(y−3)2=10\left(x-5\right)^2-\left(y-3\right)^2=10(x−5)2−(y−3)2=10
(x+5)2+(y−3)2=10\left(x+5\right)^2+\left(y-3\right)^2=10(x+5)2+(y−3)2=10
To derive the equation of a hyperbola using foci and the distance formula, the standard form should be of the type:
a2x2−b2y2=1a^2x^2-b^2y^2=1a2x2−b2y2=1
a2x2+b2y2=1a^2x^2+b^2y^2=1a2x2+b2y2=1
a2x2+b2y2=10a^2x^2+b^2y^2=10a2x2+b2y2=10
It is done.