SOLUTION: Please help me solve this equation: if the latus rectum of an ellipse is 5 and eccentricity is 2/3, find the value of a and b ?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help me solve this equation: if the latus rectum of an ellipse is 5 and eccentricity is 2/3, find the value of a and b ?      Log On


   

Question 925977: Please help me solve this equation:
if the latus rectum of an ellipse is 5 and eccentricity is 2/3, find the value of a and b ?
Answer by srinivas.g(540) About Me  (Show Source):
You can put this solution on YOUR website!
latus rectum of an ellipse =+2b%5E2%2Fa
eccentricity = +sqrt%281-b%5E2%2Fa%5E2%29
+2b%5E2%2Fa++=5
move a to the right
2b^2 = 5a
divide with 2 on both sides
+2b%5E2%2F2+=++5a%2F2
+b%5E2+=5a%2F2
put above in eccentricity equation

eccentricity = +sqrt%281-b%5E2%2Fa%5E2%29
2%2F3+=+sqrt%281-b%5E2%2Fa%5E2%29
2%2F3+=sqrt%281-%285a%2F2%29%2Fa%5E2%29
2%2F3+=sqrt%281-5%2F2a%29
squaring on both sides
+%282%2F3%29%5E2++=%28sqrt%281-5%2F2a%29%29%5E2
+4%2F9=1-5%2F2a
move 1 to the left
4%2F9-1+=-5%2F2a
+4%2F9+-9%2F9+=-5%2F2a
+%284-9%29%2F9+=-5%2F2a
-5%2F9+=-5%2F2a
5%2F9+=5%2F2a
divide with 5 on both sides
+5%2F9+%2A1%2F5+=+5%2F2a%2A1%2F5
+1%2F9+=+1%2F2a
2a =9
on dividing with 2 on both sides
a = 9/2
but +b%5E2++=+5%2Aa%2F2
=5%2A+%289%2F2%29%2F2
=45%2F4
++b+=+sqrt%2845%2F4%29