Respuesta :

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]

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