Question

Maths question
Please answer fast
It's urgent as tomorrow is exam ​

Answer 1

Hey buddy, here we go


(i)
$2 {x}^{2} + 2 {y}^{2} - 10x + 6y + 8 = 0$
Dividing both sides by 2,
${x}^{2} + {y}^{2} - 5x + 3y + 4 = 0 \\ {x}^{2} - 5x + {( \frac{5}{2} )}^{2} + {y}^{2} + 3y + { (\frac{3}{2} )}^{2} = - 4 + \frac{25}{4} + \frac{9}{4} \\ {(x - \frac{5}{2}) }^{2} + {(y - \frac{3}{2} )}^{2} = \frac{18}{4} = \frac{9}{2} = { (\frac{3}{ \sqrt{2} }) }^{2}$
Therefore,
The center is,
((5/2),(-3/2)) and radius=3/ sqrt2.


(ii)
$16 {x}^{2} + 25 {y}^{2} = 1600 \\ \frac{16 {x}^{2} }{1600} + \frac{ 25{y}^{2} }{1600} = 1 \\ \frac{ {x}^{2} }{100} + \frac{ {y}^{2} }{64} = 1$

From this equation,
we get,
$a = \sqrt{100} = 10 \\ b = \sqrt{64} = 8$
(ignoring -ve values since distances are always positive).

$c = ae = \sqrt{ {a}^{2} - {b}^{2} } \\ ae = \sqrt{100 - 64} = \sqrt{36} = 6\\ 10e = 6 \\ e = \frac{6}{10} = 0.6$


so eccentricity = 0.6


Coordinate of foci= (+6,0) and (-6,0)
latus rectum=
$\frac{2 {b}^{2} }{a} = \frac{2 \times 64}{10} = \frac{128}{10} = 12.8$






Finally OVERRR!!!