# Find the centre,vertices foci,lengths of latera recta and the eccencivity of ellipse whose equation is? 49x²+144y²-196x-720y-668=0

Given equation of ellipse

$49 {x}^{2} + 144 {y}^{2} - 196 x - 720 y - 668 = 0$

$49 \left({x}^{2} - 4 x + 4\right) + 144 \left({y}^{2} - 5 y + \frac{25}{4}\right) - 196 - 900 - 668 = 0$

$49 {\left(x - 2\right)}^{2} + 144 {\left(y - \frac{5}{2}\right)}^{2} = 1764$

$\setminus \frac{{\left(x - 2\right)}^{2}}{36} + \setminus \frac{{\left(y - \frac{5}{2}\right)}^{2}}{\frac{49}{4}} = 1$

$\setminus \frac{{\left(x - 2\right)}^{2}}{{6}^{2}} + \setminus \frac{{\left(y - \frac{5}{2}\right)}^{2}}{{\left(\frac{7}{2}\right)}^{2}} = 1$

Comparing above equation with the standard form of ellipse

${X}^{2} / {a}^{2} + {Y}^{2} / {b}^{2} = 1$ we get

$X = x - 2 , Y = y - \frac{5}{2} , a = 6 , b = \frac{7}{2}$

Eccentricity of ellipse

$e = \setminus \sqrt{1 - {b}^{2} / {a}^{2}} = \setminus \sqrt{1 - \setminus \frac{\frac{49}{4}}{36}} = \setminus \frac{\sqrt{95}}{12}$

Center of ellipse

$X = 0 , Y = 0$

$x - 2 = 0 , y - \frac{5}{2} = 0$

$\left(2 , \frac{5}{2}\right)$

Vertices of Ellipse

(X=\pm a, Y=0) \ & \ (X=0, Y=\pm b)

(x-2=\pm6, y-5/2=0)\ &\ (x-2=0, y-5/2=\pm7/2)

(2\pm6, 5/2)\ &\ (2, 5/2\pm7/2)

Focii of ellipse

$\left(X = \setminus \pm a e , Y = 0\right)$

$\left(x - 2 = \setminus \pm \setminus \frac{\sqrt{95}}{2} , y - \frac{5}{2} = 0\right)$

$\left(2 \setminus \pm \setminus \frac{\sqrt{95}}{2} , \frac{5}{2}\right)$

Latus rectum of ellipse

$\setminus \frac{2 {b}^{2}}{a} = \setminus \frac{2 \left(\frac{49}{4}\right)}{6} = \frac{49}{12}$