what is the vertices of 7x^2+5y^2=35

To find the vertices of the ellipse represented by the equation 7x^2 + 5y^2 = 35, you can follow these steps:

Step 1: Rearrange the equation.
Start by isolating the x^2 term and the y^2 term by dividing both sides of the equation by 35:
(x^2)/5 + (y^2)/7 = 1

Step 2: Determine the semi-major and semi-minor axes.
The equation in Step 1 is in the standard form of an ellipse:
(x^2)/(a^2) + (y^2)/(b^2) = 1

The semi-major axis (a) is the square root of the denominator of the x^2 term, and the semi-minor axis (b) is the square root of the denominator of the y^2 term.

In this case, a = √5 and b = √7.

Step 3: Identify the center of the ellipse.
The center of the ellipse is at the origin (0,0) since there is no (h, k) term present.

Step 4: Locate the vertices.
The vertices are the points on the major axis where the ellipse intersects the x-axis. The x-coordinate of each vertex is ±a, and the y-coordinate is 0. In this case, the vertices are (±√5, 0).

Therefore, the vertices of the ellipse 7x^2 + 5y^2 = 35 are (±√5, 0).