Answers
distance between focii=2ae
but given e=1/2
2ae=4
a=4
e=squareroot of asquare minusbsquare/asquare
bsquare=12
over 7 years ago
M.santhosh...
foci of ellipse is ae,0 so ae=2 and e=1upon 2 eccentricity a ^2-b^2upon a^2 so calcuate b value then final equation of ellipse is x^2\a^ +y^2\b^2=1
about 8 years ago
Hariharan
Given foci are (2,0) that is ae=2 where a is semi minor axis
from that we get a=4
and a^2=16
we know that b^2=a^2(1-e^2)
on solving we get b^2=12
since foci are on x-axis required equation is of the form x^2/a^2+y^2/b^2=1
about 8 years ago
Shiva Kond...
foci(+_ae'o) so ae=2.substite e value then u get a=4,b2=a2(1-e2),then we get b2=12
over 8 years ago
Jahnavi
if a is the length of semimajor axis of ellipse and e is its eccentricity then the X co ordinate of its focus is ae
over 9 years ago
K raghu
foci of ellipse is (ae,0) so ae=2 and given e=1/2 so a=4 and b=square root of 12 so the answer is second option
about 10 years ago
P.lahari
given ae =2, e=1/2
a(1/2)=2
a=4
formula: b.b=a.a(1-e.e)
b.b=16(1-1/4)
b.b=12
x.x/16+y.y/12=1
about 10 years ago
Sruthi
foci of the ellipse is (+ae,0),(-ae,0)
ae = 2 , e = 1/2
a(1/2) = 2
a = 4
in the ellipse b2 = a2(1-e2)
b2 = 16( 1- 1/4)
b2 = 12
a2 = 16, b2=12
Equation of the ellipse is x2/a2 + y2/b2 = 1
x2/16+ y2/12 = 1
ANSWER : 2
about 10 years ago
Manoj Kuma...
(ae,0)=(2,0) on comparing we get ae=2 and given that ecentricity(e)=1/2 so a.1/2=2 i.e . a=4 and
about 10 years ago
Ruchitha
focus=(+/-ae,0) and e2=1-b2/a2 (2s are in the power).e is given so using the first expression a can be found.once a is found,use the 2nd exp n find b..
n then put the values of a2 n b2 in the std eq of an ellipse i.e. x2/a2 + y2/b2=1.(2s are in power). you'll get the answer :)
about 10 years ago
Madonna
foci: ae=2
e=0.5, a=4
e=sqrt(1-(b^2/a^2))
b=2sqrt(3)
option (2) is correct
about 10 years ago
Puneet Sin...
focal distance formula =2ae by ths v get the value of a thn by eccenricity formulae u get b
about 10 years ago
Avinash
foci=(+-2,0) means c=2.
e=c/a. so a=c/e ie a=4.
b=sqrt(a^2-c^2)
b=sqrt(8)
almost 11 years ago
Praveen
foci are {(+ -)ae}. ae=2, e=0.5 >> a = 4, now b*b = a*a (1 - e*e), the equation becomes equivalent to the option 2.
about 11 years ago
Priyansh
distance between the foci is 2ae. find the distance between the points (2,0) and(-2,0)? then made it equal to 2ae then u will get the vaiue of a .find b using a and e? then construct equation of ellipse?
about 11 years ago
Sandeep
the coordinates of foci are along X-axis so major axis lies on X-axis.distance between the foci is 2ae=4 so we can get 'a' from this. also we can find out 'b' using the formula
b^2=a^2[1-e^2]
centre is origin so the equation of ellipse is of the form x^2/a^2 +y^2/b^2 =1
on substituting a,b values we can get the required equation.
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about 11 years ago
Venkatesh
ae=2 and e=1/2 then vve get a=4 and from eccentricity formula vve get value of b ans. 2
about 11 years ago
Bhavesh bh...
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