# The Eccentricity of the Ellipse 9x2 + 25y2 − 18x − 100y − 116 = 0, is - Mathematics

MCQ
Sum

The eccentricity of the ellipse 9x2 + 25y2 − 18x − 100y − 116 = 0, is

• 25/16

• 4/5

• 16/25

• 5/4

#### Solution

$\frac{4}{5}$
$9 x^2 - 18x + 25 y^2 - 100y - 116 = 0$
$\Rightarrow 9( x^2 - 2x) + 25( y^2 - 4y) = 116$
$\Rightarrow 9( x^2 - 2x + 1) + 25( y^2 - 4y + 4) = 116 + 100 + 9$
$\Rightarrow 9(x - 1 )^2 + 25(y - 2 )^2 = 225$
$\Rightarrow \frac{9(x - 1 )^2}{225} + \frac{25(y - 2 )^2}{225} = 1$
$\Rightarrow \frac{(x - 1 )^2}{25} + \frac{(y - 2 )^2}{9} = 1$
$\text{Comparing it with }\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\text{ we get: }$
$a = 5\text{ and }b = 3$
Here, a > b, so the major and the minor axes of the ellipse are along the x - axis and y - axis, respectively .
$\text{ Now,} e = \sqrt{1 - \frac{b^2}{a^2}}$
$\Rightarrow e = \sqrt{1 - \frac{9}{25}}$
$\Rightarrow e = \sqrt{\frac{16}{25}}$
$\Rightarrow e = \frac{4}{5}$

Concept: Introduction of Ellipse
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Q 20 | Page 29