Question

Is it possible for
x2/4 + y2 /9 = 1 to have foci at (−c, 0) and (c, 0) for some real number c?

Answer 1

Answer:No

Step-by-step explanation:

Given

Equation of ellipse

$\frac{x^2}{4}+\frac{y^2}{9}=1$

this is the equation of a vertical Ellipse

which is in the form of $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

and its eccentricity is given by

$e=\sqrt{1-\frac{b^2}{a^2}}$

$e=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}$

and Focii of Vertical ellipse is $(0,\pm ae)$

$ae=3\times \frac{\sqrt{5}}{3}=\sqrt{5}$

Focii are $(0,\pm \sqrt{5})$

so (-c,0) and (c,0) cannot be the focii of ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$