If the foci of an ellipse are ( ± √5, 0) and its eccentricity is (√5/3), then the equation of the ellipse is

Question

If the foci of an ellipse are ( ± √5, 0) and its eccentricity is (√5/3), then the equation of the ellipse is (A) 9 x2+4 y2=36 (B) 4 x2+9 y2=36 (C) 36 x2+9 y2=4 (D) 9 x2+36 y2=4

Tardigrade Admin · Accepted Answer

∵ a e= ± √5 ⇒ a= ± √5((3/√5))= ± 3 ⇒ a2=9 ∴ b2=a2(1-e2)=9(1-(5/9))=4 Hence, equation of ellipse (x2/9)+(y2/4)=1 ⇒ 4 x2+9 y2=36

Q. If the foci of an ellipse are $(\pm 5, 0)$ and its eccentricity is $\frac{5}{3}$ , then the equation of the ellipse is

$9 x^{2} + 4 y^{2} = 36$

$4 x^{2} + 9 y^{2} = 36$

$36 x^{2} + 9 y^{2} = 4$

$9 x^{2} + 36 y^{2} = 4$

$∵ a e = \pm 5$
$\Rightarrow a = \pm 5 (\frac{3}{5}) = \pm 3$
$\Rightarrow a^{2} = 9$
$∴ b^{2} = a^{2} (1 - e^{2}) = 9 (1 - \frac{5}{9}) = 4$
Hence, equation of ellipse
$\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$
$\Rightarrow 4 x^{2} + 9 y^{2} = 36$

Q. If the foci of an ellipse are (±5​,0) and its eccentricity is 35​​, then the equation of the ellipse is

9x2+4y2=36

4x2+9y2=36

36x2+9y2=4

9x2+36y2=4