Question

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

Answer 1

Answer:

(-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

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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

$\frac{16x^2}{400}-\frac{25y^2}{400}=\frac{400}{400}\\\frac{x^2}{25}-\frac{y^2}{16}=1$

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

$\frac{x^2}{25}=1\\x^2=25\\x=\pm 5$

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.