To answer this question we will complete 2 perfect trinomials.
Notice that:
[tex]x^2+4y^2+2x-24y-63=x^2+2x+4(y^2-6y)-63.[/tex]
Then:
[tex]x^2+2x+4(y^2-6y)-63=0.[/tex]
Compleating the perfect trinomials we get:
[tex]x^2+2x+4-4+4(y^2-6y+9-9)-63=0.[/tex]
Simplifying the above result we get:
[tex]\begin{gathered} (x+2)^2-4+4(y-3)^2-36-63=0, \\ (x+2)^2+4(y-3)^2-103=0. \end{gathered}[/tex]
Adding 103 to the above result we get:
[tex](x+2)^2+4(y-3)^2=103.[/tex]
Dividing by 103 we get:
[tex]\frac{(x+2)^2}{103}+\frac{4(y-3)^2}{103}=1.[/tex]
We can rewrite the above equation as follows:
[tex]\frac{(x+2)^2}{(\sqrt{103})^2}+\frac{(y-3)^2}{(\frac{\sqrt{103}}{2})^2}=1.[/tex]
Now, notice that the above equation is as follows:
[tex]\frac{(x-k)^2}{a^2}+\frac{(y-h)^2}{b^2}=1.[/tex]
Therefore the given equation represents an ellipse.
Answer: The given equation represents an ellipse.
[tex]\frac{(x+2)^2}{(\sqrt{103})^2}+\frac{(y-3)^2}{(\frac{\sqrt{103}}{2})^2}=1.[/tex]
