Respuesta :
Solution:
The standard form of eqution of a conic section is;
[tex]\begin{gathered} Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \\ \\ \text{ Where;} \\ A,B,C,D,E,F\text{ are real numbers;} \\ \\ A0,B\ne0,C\ne0 \end{gathered}[/tex]Given the equation of the conic section;
[tex]3x^2+5y^2-12x+30y=-42[/tex]Hence;
[tex]A=3,C=5,B=0[/tex]Then;
[tex]\begin{gathered} B^2-4AC=(0)^2-4(3)(5) \\ \\ B^2-4AC=-60 \\ \\ B^2-4AC<0 \\ \\ because\text{ }-60<0 \end{gathered}[/tex]Thus;
[tex]B^2-4AC<0,\text{ then the conic section is an ellipse}[/tex]ANSWER: The equation given is an ellipse.
The standard form of an ellipse is;
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex][tex]\begin{gathered} \frac{(x-2)^2}{(\sqrt{5})^2}+\frac{(y-(-3))^2}{(\sqrt{3})^2}=1 \\ \\ \frac{(x-2)^2}{5}+\frac{(y+3)^2}{3}=1 \end{gathered}[/tex]