Question

Find the eccentricity, foci,vertices and the length of the latus rectum of the ellipse x2+4y2+8y-2x+1=0

Answer 1

We have been given the equation of ellipse as:

$x^2+4y^2+8y-2x+1=0$

Let us try to write the equation of ellipse in the standard notation:

$x^2+4y^2+8y-2x=-1$

$(x^2-2x+1)+4(y^2+2y+1)=-1+1+4$

$\implies (x-1)^2+4(y+1)^2=4$

$\frac{(x-1)^2}{2^2} +\frac{(y+1)^2}{1^2} =1$

Shifting the origin to (1,-1) without rotating the axes. We get the new coordinates of the ellipse:

$x=X+1,y=Y-1$

Our new equation of the ellipse is:

$\frac{X^2}{2^2} +\frac{Y^2}{1^2} =1$

It is in the form of :

$\frac{X^2}{a^2} +\frac{Y^2}{1^2} =1$ where, $a>b$

On comparison with the older equation we get:

$a^2=2^2$ and $b^2=1$

$a=2, b=1$

1. Finding the eccentricity of the ellipse:

$b^2=a^2(1-e^2)$

Plugging the values of 'a' and 'b', we get:

$1=4(1-e^2)$

$e^2=1-\frac{1}{4} =\frac{3}{4}$

$e=\frac{\sqrt3}{4}$

Therefore, the eccentricity of the ellipse is $\frac{\sqrt 3}{2}$.

Coordinates of focii are:

$X=\pm ae, y=0$

$X=\pm \sqrt 3, y=0$

Coordinates of the focii with respect to the old axes is:

$(1 \pm \sqrt 3, -1)$ (since we had shifted the origin to (1, -1))

Length of Latus rectum is given by:

$\frac{2b^2}{a} =\frac{2(1)^2}{2} =1$.

Answer 2

Answer:

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Step-by-step explanation