find the equation of the ellipse whose eccentricity is 3/4 ,focuson the y axis and passing through the point (6,4)

General equation of ellipse with focus on y -axis is $\frac{y^2}{a^2}+\frac{x^2}{b^2}= 1 (a^2>b^2)$
And e = $\sqrt{1-\frac{b^2}{a^2}}$
And the ellipse is passing through (6,4)
And b² = a² (1-e² )
So putting the point ( 6,4) in general equation, we get 16/a²  + 36/b² = 1 (1)
So $$\begin{gathered} \frac{16}{a^2}+\frac{36}{a^2(1-{\displaystyle \frac{9}{16}})}=1 \\ \frac{1}{a^2}(16 + \frac{36}{\left({\displaystyle \frac{7}{16}}\right)}=1 \\ Or 16 + \frac{36\times 16}{7} = a^2 \\ \frac{688}{7}=a^2 \end{gathered}$$
And $$\begin{gathered} \frac{36}{b^2}=1-\frac{16}{{\displaystyle \frac{688}{7}}}=1-\frac{112}{688}=\frac{576}{688} \\ So b^2=\frac{36\times 688}{576} = 43 \end{gathered}$$
So the equation of the ellipse will be y² /(688/7) + x² / (43) = 1