Find length of latus rectum and vertices respectively of the ellipse 9x2 + 2 5 y2 = 225.

Question

Find length of latus rectum and vertices respectively of the ellipse 9x2 + 2 5 y2 = 225. (A) (18/5), (± 5, 0) (B) (5/18), (± 5, 0) (C) (5/6), (± 5, 2) (D) (5/18), (0, ±5)

Tardigrade Admin · Accepted Answer

The given equation can be rewritten as (x2/25)+(y2/9)=1 This shows that a = 5 and b = 3. Hence length of latus rectum =(2b2/a)=(18/5) and vertices are (5,0) and (-5,0).

Q. Find length of latus rectum and vertices respectively of the ellipse $9 x^{2} + 25 y^{2} = 225$ .

$\frac{18}{5}, (\pm 5, 0)$

$\frac{5}{18}, (\pm 5, 0)$

$\frac{5}{6}, (\pm 5, 2)$

$\frac{5}{18}, (0, \pm 5)$

The given equation can be rewritten as
$\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$
This shows that $a = 5$ and $b = 3$ . Hence length of latus rectum $= \frac{2 b ^{2}}{a} = \frac{18}{5}$ and vertices are $(5, 0)$ and $(- 5, 0)$ .

Q. Find length of latus rectum and vertices respectively of the ellipse 9x2+25y2=225.

518​,(±5,0)

185​,(±5,0)

65​,(±5,2)

185​,(0,±5)