9x2+4y2 = 36 The foci are located at:

Accepted Solution

Answer:The foci are located at [tex](0,\pm \sqrt{5})[/tex]Step-by-step explanation:The standard equation of an ellipse with a vertical major axis is [tex]\frac{x^2}{b^2}+\frac{y^2}{a^2}=1[/tex] where [tex]a^2\:>\:b^2[/tex].The given equation is [tex]9x^2+4y^2=36[/tex].To obtain the standard form, we must divide through by 36.[tex]\frac{9x^2}{36}+\frac{4y^2}{36}=\frac{36}{36}[/tex]We simplify by canceling out the common factors to obtain;[tex]\frac{x^2}{4} +\frac{y^2}{9}=1[/tex]By comparing this equation to [tex]\frac{x^2}{b^2}+\frac{y^2}{a^2}=1[/tex]We have [tex]a^2=9,b^2=4[/tex].To find the foci, we use the relation: [tex]a^2-b^2=c^2[/tex]This implies that:[tex]9-4=c^2[/tex][tex]c^2=5[/tex][tex]c=\pm\sqrt{5}[/tex]The foci are located at [tex](0,\pm c)[/tex]Therefore the foci are [tex](0,\pm \sqrt{5})[/tex]Or[tex](0,-\sqrt{5})[/tex] and [tex](0,\sqrt{5})[/tex]