Find the equation of ellipse if it passes through (3, 2) whose centre(

1.	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» $\textbf{Given:}$ $\textsf{Centre of the ellipse is (0,0) and}$ $\mathsf{eccentricity\;is\;\dfrac{\sqrt{3}}{<klux>2</klux>}}$ $\textbf{To find:}$ $\textsf{The equation of the ellipse}$ $\textbf{Solution:}$ $\textsf{Since the major axis is along y axis, the equation of ellipse}$ $\textsf{can be written as}$ $\mathsf{\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=<klux>1</klux>}$ $\textsf{It passes throug (3,2)}$ $\mathsf{\dfrac{9}{b^2}+\dfrac{4}{a^2}=1}$ ---------(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) $\textsf{Using (2) in (1)}$ $\mathsf{\dfrac{9}{\dfrac{a^2}{4}}+\dfrac{4}{a^2}=1}$ $\mathsf{\dfrac{36}{a^2}+\dfrac{4}{a^2}=1}$ $\mathsf{\dfrac{40}{a^2}=1}$ $\implies\mathsf{a^2=40}$ $\mathsf{Now,\;b^2=\dfrac{40}{4}}$ $\mathsf{Now,\;b^2=10}$ $\therefore\textsf{The equation of the ellipse is}$ $\boxed{\mathsf{\dfrac{x^2}{10}+\dfrac{y^2}{40}=1}}$