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

1 Answer

Given equation of ellipse

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

49(x^2-4x+4)+144(y^2-5y+25/4)-196-900-668=0

49(x-2)^2+144(y-5/2)^2=1764

\frac{(x-2)^2}{36}+\frac{(y-5/2)^2}{49/4}=1

\frac{(x-2)^2}{6^2}+\frac{(y-5/2)^2}{(7/2)^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-5/2, a=6, b=7/2

Eccentricity of ellipse

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

Center of ellipse

X=0, Y=0

x-2=0, y-5/2=0

(2, 5/2)

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

(X=\pmae, Y=0)

(x-2=\pm\sqrt{95}/2, y-5/2=0)

(2\pm\sqrt{95}/2, 5/2)

Latus rectum of ellipse

\frac{2b^2}{a}=\frac{2(49/4)}{6}=49/12