91Ó°ÊÓ

We are given an equation as :$9x^2+4y^2-18x+8y=23$

When either xor y is squared — not both then it is parabola.

When xand y are both squared and the coefficients are positive but different then it is ellipse

When xand yare both squared, and exactly one of the coefficients is negative and exactly one of the coefficients is positive then it is a hyperbola

So from the equation :$9x^2+4y^2-18x+8y=23$has both variable squared and has positive co-efficient so it is a ellipse

We need to change the given equation to match the equation of an ellipse :

$9x^2+4y^2-18x+8y=23$

$\frac{{(x-1)}^2}{4}+\frac{{(y+1)}^2}{9}=1$

The equation of an ellipse is given by :

$\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1or\frac{{(x-h)}^2}{b^2}+\frac{{(y-k)}^2}{a^2}$where $a>b$

The given equation : $\frac{{(x-1)}^2}{2^2}+\frac{{(y+1)}^2}{3^2}=1$is of the type $\frac{{(x-h)}^2}{b^2}+\frac{{(y-k)}^2}{a^2}=1$where $a=3,b=2,h=1andk=-1$

The center of an ellipse is : $(h,k)=(1,-1)$

The vertices of a ellipse is given by : $(h,k+a)=(1,-1Â\pm 3)$here we get : $(1,2),(1,-4)$

The foci of an ellipse is given by : $(h,kÂ\pm c)=(1,-1Â\pm \sqrt{5})$, we can find$c$by using $c^2=a^2+b^2$