Find the center, vertices and foci of 16x^2-9y^2+144=0

Question

tkhunny · Answer

Have you considered writing it in a familiar form?

$16x^{2} - 9y^{2} + 144 = 0$

$9y^{2} - 16x^{2} = 144$ -- Okay, it look like an hyperbola with center (0,0)

$\dfrac{y^{2}}{16} - \dfrac{x^{2}}{9} = 1$ -- Okay, it has a major axis of lenglt 2*4 in the y-direction and a minor axis length of 2*3 in the x-direction.

All that from just putting it in a familiar form.  Now what?

anonymous · Answer

i dont know how to get the center of it

tkhunny · Answer

This is why you should be acquainted with the FORM.

$\dfrac{(y-k)^{2}}{a^{2}} - \dfrac{(x-h)^{2}}{b^{2}} = 1$

Center is (h,k) - Every time.  We have zero (0) in both cases in your problem.