MCQ
For the ellipse $25{x^2} + 9{y^2} - 150x - 90y + 225 = 0$ the eccentricity $e = $
- A$2\over5$
- B$3\over5$
- ✓$4\over5$
- D$1\over5$
$25{x^2} + 9{y^2} - 150x - 90y + 225 = 0$
$ \Rightarrow $$25\,{(x - 3)^2} + 9{(y - 5)^2} = 225$
$ \Rightarrow $$\frac{{{{(x - 3)}^2}}}{9} + \frac{{{{(y - 5)}^2}}}{{25}}$ = 1. Here $b > a$
$\therefore $ Eccentricity $e = \sqrt {1 - \frac{{{a^2}}}{{{b^2}}}} = \sqrt {1 - \frac{9}{{25}}} $
$ = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$.
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Statement $1:$ $\left| {{Z_1} - {Z_2}} \right|\, \ge \left| {{Z_{_1}}} \right|\, - \,\left| {{Z_{_2}}} \right|$
Statement $2:$ $\left| {{Z_1} + {Z_2}} \right|\, \le \left| {{Z_{_1}}} \right|\, + \,\left| {{Z_{_2}}} \right|$
$(S1)$ : If $P ( A )=0$, then $A =\phi$
$( S 2)$ : If $P ( A )=$, then $A =\Omega$
Then