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=√1−b2a2 ⇒12=√1−b216 On squaring both sides, we get: 14=16−b216 ⇒64−4b2=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(1−e2) ⇒5a2=a2[1−(23)2]    [∵e=23] ⇒5a2=a2[1−49] ⇒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(1−e2) ⇒b2=16[1−(12)2] ⇒b2=16[1−14] ⇒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(1−e2) ⇒b2=36(1−14) ⇒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(1−e2)  ⇒b2=25(1−1625) ⇒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(1−e2) ⇒a2=169(1−25169) ⇒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(1−e2)  ⇒b2=36(1−1636) ⇒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=√1−b2a2 ⇒513=√1−b2169 On squaring both sides, we get: 25169=169−b2169 ⇒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=√1−a2b2 ⇒68=√1−a264 On squaring both sides, we get: 3664=64−a264 ⇒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=√1−b2a2 ⇒34=√1−b216 On squaring both sides, we get: 916=16−b216 ⇒b2=7 ∴x216+y27=1 This is the required equation of the ellipse.

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