For the ellipse $25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0$ , the eccentricity is equal to

$\frac{2}{5}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{3}{5}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\sqrt{\frac{15}{24}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{4}{5}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$\frac{4}{5}$

Explanation for the correct option:
Equation for an ellipse:
$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1 . . . (A)$
Eccentricity of an ellipse.
$\Rightarrow 25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0 \Rightarrow 25 (x^{2} - 6 x) + 9 (y^{2} - 10 y) + 225 = 0 \Rightarrow 25 (x^{2} - 6 x + 9) + 9 (y^{2} - 10 y + 25) = 225 \frac{{(x - 3)}^{2}}{9} + \frac{{(y - 5)}^{2}}{25} = 1 . . . (B) C o m p a r e e q^{n} (A) & (B) H e r e, a = 3, b = 5 N o w, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} e = \sqrt{1 - \frac{9}{25}} e = \frac{4}{5}$
Hence, the correct option is $(D)$

Suggest Corrections

Similar questions

Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i) $4 x^{2} + 9 y^{2} = 1$
(ii) $5 x^{2} + 4 y^{2} = 1$
(iii) $4 x^{2} + 3 y^{2} = 1$
(iv) $25 x^{2} + 16 y^{2} = 1600$
(v) $9 x^{2} + 25 y^{2} = 225$

Q. Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i) 4x² + 9y² = 1
(ii) 5x² + 4y² = 1
(iii) 4x² + 3y² = 1
(iv) 25x² + 16y² = 1600.
(v) 9x² + 25y² = 225

Q. For the ellipse

25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0,

the eccentricity

e

is equal to:

Q. Foci of the ellipse

25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0

are :

Foci of the Ellipse $25 x^{2} + 9 y^{2} - 150 x - 90 y + 225$ = 0 are

Join BYJU'S Learning Program

For the ellipse 25x2+9y2-150x-90y+225=0, the eccentricity is equal to

For the ellipse $25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0$ , the eccentricity is equal to