Consider the equation 16 x 2 + 4 y 2 + 32 x − 16 y − 32 = 0 16x^2+4y^2+32x-16y-32=0 16 x 2 + 4 y 2 + 32 x − 16 y − 32 = 0
16 ( x 2 + 2 x ) + 4 ( y 2 − 4 y ) = 32 16(x^2+2x)+4(y^2-4y)=32 16 ( x 2 + 2 x ) + 4 ( y 2 − 4 y ) = 32
16 ( x 2 + 2 x + 1 − 1 ) + 4 ( y 2 − 4 y + 4 − 4 ) = 32 16(x^2+2x+1-1)+4(y^2-4y+4-4)=32 16 ( x 2 + 2 x + 1 − 1 ) + 4 ( y 2 − 4 y + 4 − 4 ) = 32
16 ( x 2 + 2 x + 1 ) − 16 + 4 ( y 2 − 4 y + 4 ) − 16 = 32 16(x^2+2x+1)-16+4(y^2-4y+4)-16=32 16 ( x 2 + 2 x + 1 ) − 16 + 4 ( y 2 − 4 y + 4 ) − 16 = 32
16 ( x + 1 ) 2 + 4 ( y − 2 ) 2 = 32 + 16 + 16 16(x+1)^2+4(y-2)^2=32+16+16 16 ( x + 1 ) 2 + 4 ( y − 2 ) 2 = 32 + 16 + 16
16 ( x + 1 ) 2 + 4 ( y − 2 ) 2 = 64 16(x+1)^2+4(y-2)^2=64 16 ( x + 1 ) 2 + 4 ( y − 2 ) 2 = 64
Dividing by 64, we get
( x + 1 ) 2 4 + ( y − 2 ) 2 16 = 1 \frac{(x+1)^2}{4}+\frac{(y-2)^2}{16}=1 4 ( x + 1 ) 2 + 16 ( y − 2 ) 2 = 1
Observe that the above equation is of the form
( x − h ) 2 a 2 + ( y − k ) 2 b 2 = 1 \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 a 2 ( x − h ) 2 + b 2 ( y − k ) 2 = 1
So, the given equation is ellipse
Center of the ellipse = ( h , k ) = ( − 1 , 2 ) =(h,k)=(-1,2) = ( h , k ) = ( − 1 , 2 )
And a 2 = 4 , b 2 = 16 a^2=4,b^2=16 a 2 = 4 , b 2 = 16
Observe that a 2 < b 2 a^2<b^2 a 2 < b 2
To find the foci of the ellipse, find c c c using the relation c 2 = b 2 − a 2 c^2=b^2-a^2 c 2 = b 2 − a 2
c 2 = 16 − 4 = 12 c^2=16-4=12 c 2 = 16 − 4 = 12
⇒ c = ± 2 3 \Rightarrow c=\pm 2\sqrt{3} ⇒ c = ± 2 3
Foci of the ellipse are: ( h , k − c ) , ( h , k + c ) = ( − 1 , 2 − 2 3 ) , ( − 1 , 2 + 2 3 ) (h,k-\sqrt{c}),(h,k+\sqrt{c})=(-1,2-2\sqrt{3}),(-1,2+2\sqrt{3}) ( h , k − c ) , ( h , k + c ) = ( − 1 , 2 − 2 3 ) , ( − 1 , 2 + 2 3 )
Ends of the major axis are: ( h , k − b ) , ( h , k + b ) = ( − 1 , 2 − 4 ) , ( − 1 , 2 + 4 ) = ( − 1 , − 2 ) , ( − 1 , 6 ) (h,k-b),(h,k+b)=(-1,2-4),(-1,2+4)=(-1,-2),(-1,6) ( h , k − b ) , ( h , k + b ) = ( − 1 , 2 − 4 ) , ( − 1 , 2 + 4 ) = ( − 1 , − 2 ) , ( − 1 , 6 )
Ends of the minor axis are:
( h − a , k ) , ( h + a , k ) = ( − 1 − 2 , 2 ) , ( − 1 + 2 , 2 ) = ( − 3 , 2 ) , ( 1 , 2 ) (h-a,k),(h+a,k)=(-1-2,2),(-1+2,2)=(-3,2),(1,2) ( h − a , k ) , ( h + a , k ) = ( − 1 − 2 , 2 ) , ( − 1 + 2 , 2 ) = ( − 3 , 2 ) , ( 1 , 2 )
Ends of the latus rectum are:
( − a 2 b , c ) , ( a 2 b , c ) = ( − 1 , 2 3 ) , ( 1 , 2 3 ) (\frac{-a^2}{b},c),(\frac{a^2}{b},c)=(-1,2\sqrt{3}),(1,2\sqrt{3}) ( b − a 2 , c ) , ( b a 2 , c ) = ( − 1 , 2 3 ) , ( 1 , 2 3 )
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