Question

Find the equation of ellipse if it passes through (3, 2) whose centre(0, 0) Eccentricity√3/2 and the major axis is on y-axis.

Answer 1

$\textbf{Given:}$

$\textsf{Centre of the ellipse is (0,0) and}$

$\mathsf{eccentricity\;is\;\dfrac{\sqrt{3}}{2}}$

$\textbf{To find:}$

$\textsf{The equation of the ellipse}$

$\textbf{Solution:}$

$\textsf{Since the major axis is along y axis, the equation of ellipse}$

$\textsf{can be written as}$

$\mathsf{\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1}$

$\textsf{It passes throug (3,2)}$

$\mathsf{\dfrac{9}{b^2}+\dfrac{4}{a^2}=1}$---------(1)

$\mathsf{But,\;b^2=a^2(1-e^2)}$

$\mathsf{b^2=a^2\left(1-\dfrac{3}{4}\right)}$

$\mathsf{b^2=a^2\left(\dfrac{1}{4}\right)}$

$\mathsf{b^2=\dfrac{a^2}{4}}$____(2)

$\textsf{Using (2) in (1)}$

$\mathsf{\dfrac{9}{\dfrac{a^2}{4}}+\dfrac{4}{a^2}=1}$

$\mathsf{\dfrac{36}{a^2}+\dfrac{4}{a^2}=1}$

$\mathsf{\dfrac{40}{a^2}=1}$

$\implies\mathsf{a^2=40}$

$\mathsf{Now,\;b^2=\dfrac{40}{4}}$

$\mathsf{Now,\;b^2=10}$

$\therefore\textsf{The equation of the ellipse is}$

$\boxed{\mathsf{\dfrac{x^2}{10}+\dfrac{y^2}{40}=1}}$

Answer 2

$\textbf{Given:}$

$\textsf{Centre of the ellipse is (0,0) and}$

$\mathsf{eccentricity\;is\;\dfrac{\sqrt{3}}{2}}$

$\textbf{To find:}$

$\textsf{The equation of the ellipse}$

$\textbf{Solution:}$

$\textsf{Since the major axis is along y axis, the equation of ellipse}$

$\textsf{can be written as}$

$\mathsf{\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1}$

$\textsf{It passes throug (3,2)}$

$\mathsf{\dfrac{9}{b^2}+\dfrac{4}{a^2}=1}$---------(1)

$\mathsf{But,\;b^2=a^2(1-e^2)}$

$\mathsf{b^2=a^2\left(1-\dfrac{3}{4}\right)}$

$\mathsf{b^2=a^2\left(\dfrac{1}{4}\right)}$

$\mathsf{b^2=\dfrac{a^2}{4}}$____(2)

$\textsf{Using (2) in (1)}$

$\mathsf{\dfrac{9}{\dfrac{a^2}{4}}+\dfrac{4}{a^2}=1}$

$\mathsf{\dfrac{36}{a^2}+\dfrac{4}{a^2}=1}$

$\mathsf{\dfrac{40}{a^2}=1}$

$\implies\mathsf{a^2=40}$

$\mathsf{Now,\;b^2=\dfrac{40}{4}}$

$\mathsf{Now,\;b^2=10}$

$\therefore\textsf{The equation of the ellipse is}$

$\boxed{\mathsf{\dfrac\red{x^2}{10}+\dfrac\red{y^2}{40}=1}}$