Find the coordinates of the vertex and focus of the ellipse 9x Square+ 4y square=0.09

To find the coordinates of the vertex and focus of an ellipse, we need to rewrite the equation in a standard form. The standard form for an ellipse is given by:

((x - h)^2)/(a^2) + ((y - k)^2)/(b^2) = 1

Where (h, k) represents the coordinates of the center, and 'a' and 'b' represent the lengths of the major and minor axes, respectively.

Starting with the given equation, we divide both sides by 0.09 to get:

(9x^2)/(0.09) + (4y^2)/(0.09) = 1

This simplifies to:

100x^2 + 225y^2 = 1

To put it in standard form, we need to divide both sides by 100:

x^2/((1/100)) + y^2/((1/225)) = 1

Now we can see that (h, k) = (0, 0), a^2 = (1/100), and b^2 = (1/225). Therefore, 'a' = 1/10 and 'b' = 1/15.

The coordinates of the vertex can be found by adding and subtracting 'a' from the center. Thus, the coordinates of the vertex are (h + a, k) and (h - a, k), which in this case are (1/10, 0) and (-1/10, 0) respectively.

Similarly, the coordinates of the focus can be found using the formula 'c' = √((a^2) - (b^2)). In this case, 'a' is already known as 1/10 and 'b' is 1/15. Thus, c = √(1/100 - 1/225) = √(9/900 - 4/900) = √(5/900) = √(1/180).

Therefore, the coordinates of the focus are (h + c, k) and (h - c, k), which in this case are (√(1/180), 0) and (-√(1/180), 0) respectively.

So, the coordinates of the vertex are (1/10, 0) and (-1/10, 0), and the coordinates of the focus are (√(1/180), 0) and (-√(1/180), 0).