Math, asked by Sherlock5244, 1 year ago

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

Answers

Answered by FelisFelis
42

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.


Answered by rajeshcblko
0

Answer:

Namaste india

Step-by-step explanation

Similar questions