Question

Find the equation of axes, driectrix, co-ordinate of focii, centre, vertices, length of latus - rectum and eccentricity of an ellipse
$\frac{(x-3{)}^2}{25}+\frac{(y-2{)}^2}{16}=1$

Answer 1

Let x - 3 = X, y - 2 = Y so equation of ellipse becomes as $\frac{X^2}{5^2}+\frac{Y^2}{4^2}=1$

equation of major axis is Y = 0

$\Rightarrow$ y = 2

equation of minor axis is X = 0

$\Rightarrow$x = 3

centre (X = 0, Y = 0)

$\Rightarrow$ $x=3,y=2,C=(3,2)$

Length of semi-major axis $a=5$

Length of major axis $2a=10$

Length of semi-manor axis $b=4$

Length of minor axis = $2b=8$

Let 'e' be eccentricity

$\therefore b^2=a^2(1-e^2)\Rightarrow e=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{25-16}{25}}=\frac{3}{5}$

Length of latus rectum = $L{L}^{\prime }=\frac{2b^2}{a}=\frac{2\times 16}{5}=\frac{32}{5}$

Co-ordinates focii are $X=\pm ae,Y=0$

$\Rightarrow S\equiv (X=3,Y=0)$ & $S^{\prime }\equiv (X=-3,Y=0)$

$\Rightarrow S\equiv (6,2)$ & $S^{\prime }\equiv (0,2)$