Question

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If the latus rectum of an ellipse subtends 90 degree at the centre of the ellipse then the eccentricity of that ellipse is

Answer 1

Solution:-We know that ellipse is always in the form.of

x^2/a^2 +y^2/b^2=1

where

b^2=a^2(1-e^2)

where e>1

now,

If AA' will be the latus rectum

then

coordinates of O'=(ae,0)

since AA' is a straight like,so,

x coordinate will be same for both points A and A'=ae

coordinates of A=(ae,y)

coordinates of B=(ae,y')

putting the value of coordinates of A in ellipse formula we will get=

a^2e^2/a^2 +y'^2/(b^2)=1

e^2+y'^2/b^2=1

y^2/b^2=1-e^2.....i)

also,we know that in an ellipse

b^2=a^2(1-e^2)

b^2/a^2=1-e^2

putting this value in first we get

y'^2/b^2=b^2/a^2

y'^2=b^4/a^2

y=+b^2/a or -b^2/a

so,

coordinates of A=(ae,b^2/2)

coordinates of A'=(ae,-b^2/a)

now , using distance formula we get

OA={a^e^2+(b^4)/a^2}^1/2

OB={a^2e^2+{(-b^2)/a^2}^2}^1/2={2a^2e^2+(2b^4)/a^2}^1/2

since A0A' is right angled triangle ,so

(AA')^2={2a^2e^2+(2b^4)/a^2}

(AA')^2={2(ae)^2+2b^4/a^2}.....ii)

here clearly we can see from figure attached

that

(AA')=2b^2/a

(AA')^2=4b^4/a^2

equating this to ii) we get

2(ae)^2+2b^4/a^2=4b^4/a^2

2(ae)^2=2b^4/a^2

a^2×e^2=a^4{(1-e^2)}^2/a^2 (putting b^4={a^2(1-e^2)}^2)

e^2=(1-e^2)^2

let e^2 =a

a=(1-a)^2

a=1+a^2-2a

a^2-3a+1=0

a={3-+(9-4)^1/2}/2 or

since eccentricity is less than 1,so we will ignore larger root as it will be greater than 1

a=(3-√5)/2

now we know a=e^2

so,

e^2=(3-√5)/2

e={(3-√5)/2}^1/2

hence the eccentricity will be {(3-√5)/2}^1/2

Answer 2

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