The area enclosed by the ellipse `(x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1` is equal to

Answer

`a rarrp; b rarr r, s;c rarr q; drarrs`

Explanation

We have
`A=ae_(E) and a=Ae_(H)`
`"or "e_(E)e_(H)=1`
`therefore" "e_(E)+e_(H)gt2" "("Using "e_(E)+e_(H)gtsqrt(e_(E)e_(H)))`
`B^(2)=A^(2)(e_(H)^(2)-1)=a^(2)(1-e_(E)^(2))=b^(2)`
`"or "(b)/(B)=1`
Also, the angle between the asymptotes is
`2tan^(-1).(B)/(A)=(2pi)/(3)`
Also, `(B)/(A)=sqrt3or(b)/(ae_(E))=sqrt3ore_(E)^(2)=(1)/(4)`
Solving `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(a^(2)e_(E)^(2))-(y^(2))/(b^(2))=1 or(2x^(2))/(a^(2))-(y^(2))/(b^(2))=1`
Now, solve.