Question

With transverse axis parallel to the x-axis, center at (2,-2), passing through (2 + 3sqrt2, 0) and (2 + 3sqrt10, 4)

Answer 1

I assume you want the equation of a hyperbola

so we know:
(x-2)^2 /a^2 - (y+2)^2 /b^2 = 1
for point(2+3√2,0)
(3√2)^2 /a^2 - 2^2 /b^2 = 1
18/a^2 - 4/b^2 = 1
18b^2 - 4a^2 = a^2 b^2

for point(2+3√10,4)
(3√10)^2 /a^2 - 6^2/ b^2 = 1
90b^2 - 36a^2 = a^2b^2

so 90b^2 - 36a^2 = 18b^2 - 4a^2
72b^2 = 32a^
36b^2 = 16a^2 or b^2 = 16a^2 /36 = 4a^2 /9
6b = ± 4a

if 6b = 4a
b = 2a/3

sub b = 2a/3 into 18b^2 - 4a^2 = a^2b^2
18(4a^2/9) - 4a^2 = a^2(4a^2/9)
times 9
72a^2 - 36a^2 = 4a^4
4a^4 = 36a^2
divide both sides by 4a^2 , a ≠ 0
a^2 = 9
then b^2 = 4a^2 /9 = 9/9 = 1

<b>(x-2)^2 /9 - (y+2)^2 = 1</b> is the equation

Answer 2

oops, messed up in the last few lines

the end should be:

a^2 = 9
then b^2 = 4a^2 /9 = 36/9 = 4

<b>(x-2)^2 /9 - (y+2)^2/4 = 1</b> is the equation