Question

If vertices are (2,-2), (2, 4) and e =1/3
then equation of the ellipse is
​

Answer 1

$\textbf{Given:}$

$\text{Vertices A(2,-2) and A'(2,4) and e=\dfrac{1}{3} }$

$\textbf{To find:}$

$\text{Equation of the ellipse}$

$\textbf{Solution:}$

$\text{We know that,}$

$\text{Centre of the ellipse is the midpoint of its vertices}$

$\implies\text{Centre}\,C(\frac{2+2}{2},\frac{-2+4}{2})$

$\implies\textbf{Centre}\bf\,C(2,1)$

$\text{Distance between vertices}\,AA'=2a$

$\implies\sqrt{(2-2)^2+(-2-4)^2}=2a$

$\implies\sqrt{36}=2a$

$\implies\,2a=6$

$\implies\bf\,a=3$

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

$b^2=3^2(1-(\frac{1}{3})^2)$

$b^2=9(1-\frac{1}{9})$

$b^2=9(\frac{8}{9})$

$\implies\bf\,b^2=8$

$\text{From the given data, it is clear that the major axis is parallel to y axis}$

$\text{The equation of the ellipse is}$

$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$

$\implies\dfrac{(x-2)^2}{8}+\dfrac{(y-1)^2}{9}=1$

$\textbf{Answer:}$

$\textbf{The equation of the ellipse is}$

$\boxed{\bf\dfrac{(x-2)^2}{8}+\dfrac{(y-1)^2}{9}=1}$

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The eccentricity of an ellipse with its centre at the origin is 1/2.if one of the directrix is x=4,then equation of ellipse is

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