What are vertices of the conic 16x² - 25y² = 400 ?

(-5, 0) and (5, 0)

Step-by-step explanation:

This equation fits the form for a hyperbola with x-intercepts.  The standard form for such an equation is

To get the equation in the question into this standard form, divide each term by 400.

To find the x-intercepts, make y = 0.

The vertices are located at the points (-5, 0) and (5, 0).

Note: There are no y-intercepts; making x = 0 produces no real solutions for y.

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