Answer:No
Step-by-step explanation:
Given
Equation of ellipse
[tex]\frac{x^2}{4}+\frac{y^2}{9}=1[/tex]
this is the equation of a vertical Ellipse
which is in the form of [tex]\frac{x^2}{b^2}+\frac{y^2}{a^2}=1[/tex]
and its eccentricity is given by
[tex]e=\sqrt{1-\frac{b^2}{a^2}}[/tex]
[tex]e=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}[/tex]
and Focii of Vertical ellipse is [tex](0,\pm ae)[/tex]
[tex]ae=3\times \frac{\sqrt{5}}{3}=\sqrt{5}[/tex]
Focii are [tex](0,\pm \sqrt{5})[/tex]
so (-c,0) and (c,0) cannot be the focii of ellipse [tex]\frac{x^2}{4}+\frac{y^2}{9}=1[/tex]