The equation of the conic section is:
[tex]x^2+9y^2+10x-18y+25=0[/tex]
Square terms are positive and have different coefficients. Then, we can say it is an ellipse.
To find the vertex form, we need to group x and y:
[tex](x^2+10x)+(9y^2-18y)+25=0[/tex]
The grouped terms of y can be factored:
[tex](x^2+10x)+9(y^2-2y)+25=0[/tex]
It will be convenient to group the 25 with the group of x, since we can have a perfect square trinomial (since the square root of 25 is 5, which is half ten):
[tex](x^2+10x+25)+9(y^2-2y)=0[/tex]
We can factor the first goup:
[tex](x+5)^2+9(y^2-2y)=0^{}[/tex]
Now, we can try to make another perfect square trinomila from the second term. We need a number whose square root multiplied by 2 gives us 2 (the coefficient of y) This number will be one. Then, we can sum 1 inside the parenthesis, but we need to substract 9 outside it in order not to alter the equation. (it is 9 because we added 1 inside a parenthesis that is multiplied by 9).
[tex](x+5)^2+9(y^2-2y+1)-9=0^{}[/tex]
Now, we can factor the second group of terms and reorganize:
[tex](x+5)^2+9(y-1)^2=9^{}[/tex]
Now, to get the equation in the vertex form, we need a 1 in the right side of the equation. Then, we can divide everything (both sides) by 9:
[tex]\frac{(x+5)^2}{9}+(y-1)^2=1[/tex]
The equation of the ellipse is now on its vertex form. Since the number dividing the x term (9) is higher than the number dividing the y term (1) we can say that the direction of the ellipse is horizontal. It´s longer axis is parallel to the x axis of the plane.
