SOLUTION: sketch and find the equation
Foci=(2,6) and (2,-6)
difference in focal radii is 4
that is all the information that it gives me. and i have to sketch and find the equation usi
Algebra ->
Quadratic-relations-and-conic-sections
-> SOLUTION: sketch and find the equation
Foci=(2,6) and (2,-6)
difference in focal radii is 4
that is all the information that it gives me. and i have to sketch and find the equation usi
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Question 35524: sketch and find the equation
Foci=(2,6) and (2,-6)
difference in focal radii is 4
that is all the information that it gives me. and i have to sketch and find the equation using that information Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! WE HAVE TO ASCERTAIN THE FOLLOWING FIRST
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Focal Radius
This term has distinctly different definitions for different authors.
Usage 1: For some authors, this refers to the distance from the center to the focus for either an ellipse or a hyperbola. This definition of focal radius is usually written c.
Usage 2: For other authors, focal radius refers to the distance from a point on a conic section to a focus. In this case the focal radius varies depending where the point is on the curve (unless the conic in question is a circle). If there are two foci then there are two focal radii.
Note: Using this second definition, the sum of the focal radii of an ellipse is a constant. It is the same as the length of the major diameter. The difference of the focal radii of a hyperbola is a constant. It is the distance between the vertices.
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NOW ON TO YOUR PROBLEM
sketch and find the equation
Foci=(2,6) and (2,-6)
difference in focal radii is 4
that is all the information that it gives me. and i have to sketch and find the equation using that information
SINCE THE DIFFERENCE IN FOCAL RADII IS GIVEN TO BE CONSTANT LET US TAKE IT AS HYPERBOLA BY THE SECOND DEFINITION.
IN WHICH CASE DIFFERENCE IN FOCAL RADII=2B=4...B=2
FOR THE STD.EQN.OF HYPERBOLA WE HAVE
(Y-K)^2/B^2-(X-H)^2/A^2=1....WE HAVE
FOCI ARE {H,K+BE} AND {H,K-BE}....H=2....K+BE=6...K-BE=-6...HENCE K=0 AND BE=6
SINCE B=2...E=6/2=3
WHERE E IS
ECCENTRICITY
=SQRT{(A^2+B^2)/B^2}=3...SQUARING
9=(A^2+4)/4
36=A^2+4
A^2=36-4=32
HENCE EQN.OF HYPERBOLA IS
Y^2/4- (X-2)^2/32=1
GRAPH IS
