Question 21537: Find the foci of the given ellipse.
25x2+9y2+150x-90y+225=0
x2=xsquared(25xsquared)
y2=ysquared(ysquared)
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 25x2+9y2+150x-90y+225=0
{(5X)^2+2*5X*15+15^2}-15^2+{(3Y)^2-2*3Y*15+15^2}-15^2+225=0
(5X+15)^2+(3Y-15)^2=225
25(X+3)^2+9(Y-5)^2=225.....DIVIDING THROUGHOUT WITH 225
(X+3)^2/(225/25)+(Y-5)^2/(225/9)=1...COMPARING WITH STANDARD EQUATION OF ELLIPSE
((X-H)^2)/A^2)+((Y-K)^2)/B^2)=1,WE GET
A=SQRT(225/25)=15/9=5/3....B=SQRT(225/9)=15/3=5...
CENTRE IS (H,K)=(-3,5)
AND ECCENTRICITY = SQRT{(B^2-A^2)/B^2)}=SQRT((5^2-(5/3)^2)/5^2)
=SQRT(8/9).........
HENCE FOCI ARE (H,K+B*E) AND (H,K-B*E)....THAT IS
{-3,5+5SQRT(8/9)}AND {-3,5-5SQRT(8/9)}
|
|
|
