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Question

Find the equation of the ellipse in the following cases:
(i) eccentricity e=12 and foci (±2,0) 
(ii) eccentricity e=23 snd length of latus-rectum=5 
(iii) eccentricity e=12 and semi-major axis =4 
(iv) eccentricity e=12 and major axis =12 
(v) The ellipse passes through (1,4) and (-6,1).
(vi) Vertices (±5,0), foci (±4,0)
(vii) Vertices (0,±13), foci (0,±5)
(viii) Vertices (±6,0), foci (±4,0)
(ix) Ends of major axis (±3,0), ends of minor axis (0,±2)
(x) Ends of major axis (0,±5), ends of minor axis (±1,0)
(xi) Length of major axis 26, foci (±5,0)
(xii) Length of minor axis 16, foci (0,±6)
(xiii) Foci (±3,0), a=4 


Solution

(i) e=12 and foci =(±2,0)
Coordinate of the foci =(±ae,0)
We have ae=2
a×12=2
a=4
Now, e=1b2a2
12=1b216
On squaring both sides, we get:
14=16b216
644b2=16
b2=484
b2=12
Now, x2a2+y2b2=1
x216+y212=1
3x2+4y248=1
3x2+4y2=48
This is the required equation of the ellipse.
(ii) Let the equation of the required ellipse be x2a2+y2b2=1    ...(i) 
The length of latus-rectum = 5
2b2a=5   ...(ii)
Now, b2=a2(1e2)
5a2=a2[1(23)2]    [e=23]
5a2=a2[149]
52=a(59)
52×59=a
a=92
a=92
a2=814
Putting a=92 in b2=5a2, we get
b2=52×92
b2=454 
substituting a2=814 and b2=454 in equation (i), we get
x2814+y2454=1
4x281+4y245=1
4x2×5+4y2×9405=1
20x2+36y2=405
This is the equation of the required ellipse.
(iii) Let the equation of the required ellipse be 
x2a2+y2b2=1      ...(i)
Then, semi-major axis = a
a=4  [semi-major axis=4]
a2=16
Now, 
b2=a2(1e2)
b2=16[1(12)2]
b2=16[114]
b2=16×34
b2=12
substituting the value of a2 and b2 in (i), we get
x216=y212=1
3x2+4y2=48
This is the required equation of the ellipse.
(iv) Let the equation of the required ellipse be x2a2+y2b2=1 where major axis =2a       ...(i)
Now, 
2a=12    [Major-axis=12]
a=6
a2=36
Now, 
b2=a2(1e2)
b2=36(114)
b2=36×34
b2=27
Substituting the value of a2 and b2 in (i), we get
x236=y227=1
19[x24+y23]=1
3x2+4y212=9
3x2+4y2=108
This is the equation of the required ellipse.
(v) The ellipse passes through (1,4) and (-6,1) 
x2a2+y2b2=1
1a2+16b2=1
1a2=α and 1b2=β
Then α+16β=1     ...(1)
It also passes through (-6,1)
x2a2+y2b2=1
36a2+1b2=1
36α+β=1     ...(2)
Solving eqs. (1) and (2), we get:
α=3115 and β=7115
Substituting the values, we get:
3x2115+7y2115=1
3x2+7y2115=1
3x2+7y2=115
This is the required equation of ellipse.
(vi) Vertices (±5,0) and focus (±4,0)
The coordinate of its vertices and foci are (±a,0) and (±ae,0), respectively.
i.e., a=5 and ae=4
e=45
Now, b2=a2(1e2) 
b2=25(11625)
b2=9
x225+y29=1
This is the required equation of the ellipse.
(vii) Vertices (0,±13), foci (0,±5)
The coordinates of its vertices and foci are (0,±b) and (0,±be), respectively
i.e. b=13 and be=5
e=513
Now, a2=b2(1e2)
a2=169(125169)
a2=144
x2144+y2169=1
This is the required equation of the ellipse.
(viii) Vertices (±6,0), foci (±4,0)
The coordinates of its vertices and foci are (±a,0) and (±ae,0), respectively
i.e. a=6 and ae=4
e=46 or 23  
Now, b2=a2(1e2) 
b2=36(11636)
b2=20
x236+y220=1
This is the required equation of the ellipse. 
(ix) Ends of major axis (±3,0), ends of minor axis (0,±2)
Let the equation of the ellipse be x2a2+y2b2=1
Ends of major axis = (±3,0)
Ends of minor axis = (0,±2)
But the coordinates of the end points of the major and the minor axes are (±a,0) and (0,±b)
a=3 and b=2
Then, x29+y24=1
This is the required equation of the ellipse.
(x) Ends of major axis (0,±5), ends of minor axis (±1,0)
Let the equation of the ellipse be x2a2+y2b2=1
Ends of major axis = (0,±5)
Ends of minor axis = (±1,0)
But the coordinates of the end points of the major and the minor axes are (±a,0) and (0,±b)
a=1andb=5
Then, x21+y25=1
This is the required equation of the ellipse.
(xi) Length of major axis 26, foci (±5,0)
Length of major axis = 26
Foci = (±5,0)
We have 2a=26
Also, ae=5
e=513
Now, e=1b2a2
513=1b2169
On squaring both sides, we get:
25169=169b2169
b2=144
Now, x2a2+y2b2=1
x2169+y2144=1
This is the required equation of the ellipse.
(xii) Length of minor axis 16 and foci (0,±6) i.e., 2b=16
b=8
and 
be=6
e=68
Now, 
e=1a2b2
68=1a264
On squaring both sides, we get:
3664=64a264
a2=28
x264+y228=1
This is the required equation of the ellipse.
(xiii) Foci (±3,0) and a=4
i.e., ae=3
e=34
Now, e=1b2a2
34=1b216
On squaring both sides, we get:
916=16b216
b2=7
x216+y27=1
This is the required equation of the ellipse. 

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