Question

Convert into standard form. Find any centres, vertices, foci, asymptotes, directices and axes of symmetry.

64x^2-36y^2-512x-216y=-124

I found the centre (4,-3), vertices (8.4,-3) , (-0.4,-3), foci (0.1,-4), (7.9,-3).

I'm having a hard time finding the asympyotes,directrices and axes of symmetry if there is any.

Answer 1

64x^2-36y^2-512x-216y = -124
64(x^2-8x)-36(y^2-6y) = -124
64(x^2-8x+16) - 36(y^2-6y+9) = -124 + 64*16 - 36*9
64(x-4)^2 - 36(y-3)^2 = 576
(x-4)^2/36 - (y-3)^2/9 = 1

Now it should be clear that you have
a=3 b=4 c=5
giving you the properties
center: (4,3)
so, the axes of symmetry are: x=4 and y=3
foci: (4±5,3)
vertices: (4±3,3)
asymptotes: 3(y-3) = ±4(x-4)
directrices: (x-4) = a^2/c = ±9/5
semi-latus rectum: b^2/a = 16/3

Answer 2

Thank you so much !

Answer 3

I assume you caught my typo:

(x-4)^2/9 - (y-3)^2/16 = 1