how to find foci of ellipse
FINDING INFORMATION FROM EQUATION OF ELLIPSE
Question :
Find the
(i) Center
(ii) Foci
(iii) Eccentricity
(iv) Equation of latus rectum
(v) Equation of directrix
(vi) Length of latus rectum
1) 9x2+4y2= 36
2) 25x2+9y2-150x-90y+225 = 0
3)3x2+4y2-12x-8y+4 = 0
4) 9 x 2 +25y 2 -18x-100y-116 = 0
1) Answer :
Equation of ellipse :
9x 2 + 4y 2 = 36
(x 2/4)+ (y 2/9)= 1
a2 = 9 and b2 = 4
a = 3 and b = 2
Since b > a, the ellipse symmetric about y-axis.
e = √1 - (4/9)
e = √( 5/9)
e = √5/3
ae = 3( √5/3)
ae = √5
a/e = 9/ √5
Center : (0, 0)
Focus : F1 (0, √5 ) and F 2 (0, - √5 )
Eccentricity : e = √5/3
Equation of latus rectum : y = ± √5
Equation of directrix : y = ± 9/ √5
Length of latus rectum :
= 2(4)/3
= 8/3
2) Answer :
Equation of ellipse :
25x2+9y2-150x-90y+225 = 0
25x 2 -150x +9y 2 -90y+225 = 0
25( x 2 -6x) + 9(y 2 -10y) + 225 = 0
25[(x-3)2-32] + 9[(y-5)2-52] + 225 = 0
25[(x-3)2-9] + 9[(y-5)2-25] + 225 = 0
25(x-3)2-225 + 9(y-5)2-225 + 225 = 0
25(x-3)2+ 9(y-5)2= 225
(x-3)2/9 + (y-5)2/25 = 1
Let X = x-3 and Y = y-5
x = X + 3 and y = Y + 5
The ellipse is symmetric about y-axis.
a2 = 25, b2 = 9
a = 5 and b = 3
e = √(1-9/25)
e = 4/5
ae = 4
a/e = 25/4
Length of latus rectum :
= 2(9)/5
= 18/5
| Referred to X an Y | Referred to x an y x = X+3, y = Y+5 | |
| Center Foci Vertices Equation of latus rectum Equation of directrix | C(0, 0) F1(0, 4) F2(0, -4) A(0, 5) A'(0, -5) Y = 4 Y = -4 Y = 25/4 Y = -25/4 | C(3, -5) F1(3, 1) F2(3, 9) A(3, 10) A'(3, 0) y = 9 y = 1 y = 45/4 y = -5/4 |
3) Answer :
Equation of ellipse :
3x 2 +4y 2 -12x-8y+4 = 0
3x2-12x+4y2-8y+4 = 0
3(x2-4x) + 4(y2-2y) + 4 = 0
3[(x-2)2-22] + 4[(y-1)2-12] + 4 = 0
3[(x-2)2-4] + 4[(y-1)2-1] + 4 = 0
3(x-2)2-12 + 4(y-1)2-4 + 4 = 0
3(x-2)2+ 4(y-1)2= 12
(x-2)2/4 + (y-1)2/3 = 1
Let X = x-2 and Y = y-1
x = X + 2 and y = Y + 1
The ellipse is symmetric about x-axis.
a2 = 4, b2 = 3
a = 2 and b =√3
e =√(1-3/4)
e = 1/2
ae = 1
a/e = 4
Length of latus rectum :
= 2(3)/2
= 3
| Referred to X an Y | Referred to x an y | |
| Center Foci Vertices Equation of latus rectum Equation of directrix | C(0, 0) F1(1, 0) F2(-1, 0) A(2, 0) A'(-2, 0) X = 1 X = -1 X = 4 X = -4 | C(2, 1) F1(3, 1) F2(1, 1) A(4, 1) A'(0, 1) x = 3 x = 1 x = 6 x = -2 |
4) Answer :
Equation of ellipse :
9 x 2 +25y 2 -18x-100y-116 = 0
9x2-18x+25y2-100y-116 = 0
9(x2-2x) + 25(y2-4y) - 116 = 0
9[(x-1)2-12] + 25[(y-2)2-22] - 116 = 0
9[(x-1)2-1] + 25[(y-2)2-4] - 116 = 0
9(x-1)2-9 + 25(y-2)2-100 -116 = 0
9(x-1)2+ 25(y-2)2= 225
(x-1)2/25 + (y-2)2/9 = 1
Let X = x-1 and Y = y-2
x = X + 1 and y = Y + 2
The ellipse is symmetric about x-axis.
a2 = 25, b2 = 9
a = 5 and b = 3
e =√(1-9/25)
e = 4/5
ae = 4
a/e = 25/4
Length of latus rectum :
= 2(9)/5
= 18/5
| Referred to X an Y | Referred to x an y | |
| Center Foci Vertices Equation of latus rectum Equation of directrix | C(0, 0) F1(4, 0) F2(-4, 0) F1(5, 0) F2(-5, 0) X = 4 X = -4 X = 25/4 X = -25/4 | C(1, 2) F1(5, 2) F2(-3, 2) A'(6, 2) A'(-4, 2) x = 5 x = -3 x = 49/4 x = -21/4 |
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how to find foci of ellipse
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