The equation of ellipse whose distance between the foci is equal to 8 and distance between the directrix is 18 , is

Question

The equation of ellipse whose distance between the foci is equal to 8 and distance between the directrix is 18 , is (A) 5 x2-9 y2=180 (B) 9 x2-5 y2=180 (C) x2+9 y2=180 (D) 5 x2+9 y2=180

Tardigrade Admin · Accepted Answer

According to given condition, 2 a e=8, (2 a/e)=18 ⇒ a=√4 × 9=6 ⇒ e=(2/3), b=6 √1-(4/9)=(6/3) √5=2 √5 Hence, the required equation is (x2/36)+(y2/20)=1 i.e., 5 x2+9 y2=180

Q. The equation of ellipse whose distance between the foci is equal to 8 and distance between the directrix is 18 , is

$5 x^{2} - 9 y^{2} = 180$

$9 x^{2} - 5 y^{2} = 180$

$x^{2} + 9 y^{2} = 180$

$5 x^{2} + 9 y^{2} = 180$

According to given condition, $2 a e = 8, \frac{2 a}{e} = 18$
$\Rightarrow a = 4 \times 9 = 6$
$\Rightarrow e = \frac{2}{3}, b = 6 1 - \frac{4}{9} = \frac{6}{3} 5 = 25$
Hence, the required equation is $\frac{x ^{2}}{36} + \frac{y ^{2}}{20} = 1$
i.e., $5 x^{2} + 9 y^{2} = 180$

Q. The equation of ellipse whose distance between the foci is equal to 8 and distance between the directrix is 18 , is

5x2−9y2=180

9x2−5y2=180

x2+9y2=180

5x2+9y2=180

According to given condition, 2ae=8,e2a​=18 ⇒a=4×9​=6 ⇒e=32​,b=61−94​​=36​5​=25​ Hence, the required equation is 36x2​+20y2​=1 i.e., 5x2+9y2=180

$5 x^{2} - 9 y^{2} = 180$

$9 x^{2} - 5 y^{2} = 180$

$x^{2} + 9 y^{2} = 180$

$5 x^{2} + 9 y^{2} = 180$

According to given condition, $2 a e = 8, \frac{2 a}{e} = 18$
$\Rightarrow a = 4 \times 9 = 6$
$\Rightarrow e = \frac{2}{3}, b = 6 1 - \frac{4}{9} = \frac{6}{3} 5 = 25$
Hence, the required equation is $\frac{x ^{2}}{36} + \frac{y ^{2}}{20} = 1$
i.e., $5 x^{2} + 9 y^{2} = 180$