SOLUTION: Here'e the question that I can't fiqure out. I might be using the wrong equation or system of equations? A Semielliptical window has a height of 28 in and a length of 64 inches.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Here'e the question that I can't fiqure out. I might be using the wrong equation or system of equations? A Semielliptical window has a height of 28 in and a length of 64 inches.      Log On


   

Question 90858: Here'e the question that I can't fiqure out. I might be using the wrong equation or system of equations?
A Semielliptical window has a height of 28 in and a length of 64 inches. Find the height of the window 25 inches from one end.
Thanks
Answer by venugopalramana(3286) About Me  (Show Source):
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Here'e the question that I can't fiqure out. I might be using the wrong equation or system of equations?
A Semielliptical window has a height of 28 in and a length of 64 inches. Find the height of the window 25 inches from one end.
Thanks
EQN.OF ELLIPSE WITH CENTER AS O(0,0), MAJOR AXIS=POQ =2A AND MINOR AXIS=ROS=2B IS
[X^2/A^2] + [Y^2/B^2]=1
WE HAVE POQ = 2A=64....PO=OQ=A=32
AND RO=OS=B=28
HENCE EQN. IS
(X^2/32^2) + (Y^2/28^2) = 1
POINT P' IS 28" FROM ONE END MEANS PP'=28
OP'=32-28=4"=X FROM CENTER.
HENCE WE GET
4^2/32^2 + (Y^2/28^2)=1
Y^2/28^2 = 1-16/32^2=1-(1/64)= 63/64
Y^2=28^2(63/64)
Y=28*SQRT(63/64) IS ALMOST EQUAL TO 28"