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Question 27474: how do you fing the focus of a parabola when it is 9.5 inches deep and 10.5 inches wide?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! how do you fing the focus of a parabola when it is 9.5 inches deep and 10.5 inches wide?WHAT DO YOU MEAN BY DEEP AND WIDE??
DO YOU MEAN THAT FOCUS IS AT 9.5 INCHES FROM VERTEX
IN THAT CASE Y^2=2*9.5*X=19X AS PER STANDARD FORMULA
AND LATUS RECTUM IS 10.5 INCHES??
IN THAT CASE Y^2=10.5*X=10.5X AS PER STANDARD FORMULA
IF YOU MEAN THAT DEPTH AND WIDTH REFER TO THOSE AT SOME ARBITRARY POINT ON THE AXIS OF PARABOLA...THEN IF THAT POINT IS AT DISTANCE P FROM FOCUS TOWARDS VERTEX.
LET ITS EQN.BE Y^2=4AX
LET COORDINATES OF ENDS OF DIAMETER AT P BE (AT^2,2AT) AND (AT^2,-2AT) WHICH SATISFY THE EQN.OF PARABOLA...
HENCE WIDTH OR DIAMETER =10.5=SQRT(0+16A^2*T^2)=4AT...................I
DEPTH OR DISTANCE FROM FOCUS IS AT^2=9.5...........................II
FROM EQN.I..T=10.5/4A=2.625/A.SUBSTITUTING IN EQN.II,WE GET
A*2.625^2/A^2=9.25
2.625^2/A=9.25
A=2.625^2/9.5
A=0.725..HENCE EQN OF PARABOLA IS
Y^2=4*0.725X=2.9X.
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