Question

The given ellipse is ( frac{(x-3)^{2}}{25}+frac{(y-2)^{2}}{16}=1 )
Let ( x-3=x, y-2=Y, ) so equation of ellipse becomes as ( frac{x^{2}}{5^{2}}+frac{y^{2}}{4^{2}} )
equation of major axis is ( Y=0 ) ( Rightarrow quad y=2 )
equation of minor axis is ( X=0 quad Rightarrow=3 ). centre ( (X=0, Y=0) )
( Rightarrow quad x=3, y=2 )
[
C equiv(3,2)
]
Length of semi-major axis a ( =5 ) Length of major axis ( 2 a=10 ) Length of semi-minor axis ( b=4 ) Length of minor axis ( =2 b=8 ) Let 'e' be eccentricity
( therefore )
( b^{2}=a^{2}left(1-e^{2}right) )
( e=sqrt{frac{a^{2}-b^{2}}{a^{2}}}=sqrt{frac{25-16}{25}}=frac{3}{5} )
Length of latus rectum ( =L L^{prime}=frac{2 b^{2}}{a}=frac{2 times 16}{5}=frac{32}{5} )
Co-ordinates focii are ( X=pm a e, Y=0 )
( Rightarrow quad S equiv(X=3, Y=0) )
( &^{prime} equiv(X=-3, Y=0) )
( S equiv(6,2) )
( &^{prime} equiv(0,2) )

# Pramole # 10 : Find the equation of axes, directrix, co-ordinate of focii, centre, vertices, length of latus - rectum and eccentricity of an ellipse 16x2 + 25y2 - 96x - 100 y + 156 = 0.

Solution