Aloha :)
$$x^2+16y^2-\frac{8}{\sqrt5}x-\frac{224}{\sqrt5}y+144=0\quad\bigg|-144$$$$\left(x^2-\frac{8}{\sqrt5}x\right)+\left(16y^2-\frac{224}{\sqrt5}y\right)=-144\quad\bigg|16\text{ ausklammern}$$
$$\left(x^2-\frac{8}{\sqrt5}x\right)+16\left(y^2-\frac{14}{\sqrt5}y\right)=-144\quad\bigg|+\pink{\frac{4^2}{(\sqrt5)^2}}+16\cdot\pink{\frac{7^2}{(\sqrt5)^2}}=\pink{160}$$
$$\left(x^2-\frac{8}{\sqrt5}x+\pink{\frac{4^2}{(\sqrt5)^2}}\right)+16\left(y^2-\frac{14}{\sqrt5}y+\pink{\frac{7^2}{(\sqrt5)^2}}\right)=-144+\pink{160}\quad\bigg|\text{bin. Formeln}$$$$\left(x-\frac{4}{\sqrt5}\right)^2+16\left(y-\frac{7}{\sqrt5}\right)^2=16\quad\bigg|\div16$$$$\frac{\left(x-\frac{4}{\sqrt5}\right)^2}{4^2}+\frac{\left(y-\frac{7}{\sqrt5}\right)^2}{1^2}=1$$
Es handelt sich also um eine Ellipse mit Mittelpunkt \(M(\frac{4}{\sqrt5}\big|\frac{7}{\sqrt5})\).
Die große Halbachse beträgt \(a=4\) und die kleine Halbachse ist \(b=1\).
